Quadratic Formula Worksheets - Math Monks - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Part 1: Solve by Factoring

1) $3x^2 + 14x - 49 = 0$
* We need two numbers that multiply to $3 \times -49 = -147$ and add to $14$.
* The numbers are $21$ and $-7$.
* Rewrite the middle term: $3x^2 + 21x - 7x - 49 = 0$.
* Factor by grouping: $3x(x + 7) - 7(x + 7) = 0$.
* $(3x - 7)(x + 7) = 0$.
* Set each factor to zero:
* $3x - 7 = 0 \rightarrow 3x = 7 \rightarrow x = \frac{7}{3}$
* $x + 7 = 0 \rightarrow x = -7$

2) $10p^2 - 26p + 12 = 0$
* First, divide everything by 2 to make it simpler: $5p^2 - 13p + 6 = 0$.
* We need two numbers that multiply to $5 \times 6 = 30$ and add to $-13$.
* The numbers are $-10$ and $-3$.
* Rewrite the middle term: $5p^2 - 10p - 3p + 6 = 0$.
* Factor by grouping: $5p(p - 2) - 3(p - 2) = 0$.
* $(5p - 3)(p - 2) = 0$.
* Set each factor to zero:
* $5p - 3 = 0 \rightarrow 5p = 3 \rightarrow p = \frac{3}{5}$
* $p - 2 = 0 \rightarrow p = 2$

3) $l^2 - 7l - 8 = 0$
* We need two numbers that multiply to $-8$ and add to $-7$.
* The numbers are $-8$ and $1$.
* Factors: $(l - 8)(l + 1) = 0$.
* Set each factor to zero:
* $l - 8 = 0 \rightarrow l = 8$
* $l + 1 = 0 \rightarrow l = -1$

4) $8t^2 + 64t + 120 = 0$
* First, divide everything by 8: $t^2 + 8t + 15 = 0$.
* We need two numbers that multiply to $15$ and add to $8$.
* The numbers are $5$ and $3$.
* Factors: $(t + 5)(t + 3) = 0$.
* Set each factor to zero:
* $t + 5 = 0 \rightarrow t = -5$
* $t + 3 = 0 \rightarrow t = -3$

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Part 2: Solve by Completing the Square

5) $5s^2 + 6s - 59 = 0$
* Move the constant to the right: $5s^2 + 6s = 59$.
* Divide by the leading coefficient (5): $s^2 + \frac{6}{5}s = \frac{59}{5}$.
* Take half of the middle coefficient ($\frac{6}{5}$), which is $\frac{3}{5}$, and square it: $(\frac{3}{5})^2 = \frac{9}{25}$.
* Add $\frac{9}{25}$ to both sides:
$$s^2 + \frac{6}{5}s + \frac{9}{25} = \frac{59}{5} + \frac{9}{25}$$
* Convert $\frac{59}{5}$ to $\frac{295}{25}$ so we can add:
$$\frac{295}{25} + \frac{9}{25} = \frac{304}{25}$$
* Write the left side as a perfect square: $(s + \frac{3}{5})^2 = \frac{304}{25}$.
* Take the square root of both sides:
$$s + \frac{3}{5} = \pm\sqrt{\frac{304}{25}}$$
$$s + \frac{3}{5} = \pm\frac{\sqrt{16 \cdot 19}}{5} = \pm\frac{4\sqrt{19}}{5}$$
* Subtract $\frac{3}{5}$ from both sides:
$$s = \frac{-3 \pm 4\sqrt{19}}{5}$$

6) $r^2 - 4r - 98 = 0$
* Move the constant to the right: $r^2 - 4r = 98$.
* Take half of the middle coefficient ($-4$), which is $-2$, and square it: $(-2)^2 = 4$.
* Add 4 to both sides:
$$r^2 - 4r + 4 = 98 + 4$$
$$(r - 2)^2 = 102$$
* Take the square root of both sides:
$$r - 2 = \pm\sqrt{102}$$
* Add 2 to both sides:
$$r = 2 \pm \sqrt{102}$$

7) $3v^2 - 6v - 34 = 0$
* Move the constant to the right: $3v^2 - 6v = 34$.
* Divide by the leading coefficient (3): $v^2 - 2v = \frac{34}{3}$.
* Take half of the middle coefficient ($-2$), which is $-1$, and square it: $(-1)^2 = 1$.
* Add 1 to both sides:
$$v^2 - 2v + 1 = \frac{34}{3} + 1$$
$$(v - 1)^2 = \frac{34}{3} + \frac{3}{3} = \frac{37}{3}$$
* Take the square root of both sides:
$$v - 1 = \pm\sqrt{\frac{37}{3}}$$
* Rationalize the denominator inside the root ($\sqrt{\frac{37}{3}} = \frac{\sqrt{37}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{111}}{3}$):
$$v - 1 = \pm\frac{\sqrt{111}}{3}$$
* Add 1 to both sides (convert 1 to $\frac{3}{3}$):
$$v = \frac{3 \pm \sqrt{111}}{3}$$

8) $9y^2 - 12y + 13 = 0$
* Move the constant to the right: $9y^2 - 12y = -13$.
* Divide by the leading coefficient (9): $y^2 - \frac{12}{9}y = -\frac{13}{9}$.
* Simplify the fraction: $y^2 - \frac{4}{3}y = -\frac{13}{9}$.
* Take half of the middle coefficient ($-\frac{4}{3}$), which is $-\frac{2}{3}$, and square it: $(-\frac{2}{3})^2 = \frac{4}{9}$.
* Add $\frac{4}{9}$ to both sides:
$$y^2 - \frac{4}{3}y + \frac{4}{9} = -\frac{13}{9} + \frac{4}{9}$$
$$(y - \frac{2}{3})^2 = -\frac{9}{9}$$
$$(y - \frac{2}{3})^2 = -1$$
* Take the square root of both sides. Since we have a negative number, the solution involves imaginary numbers ($i$):
$$y - \frac{2}{3} = \pm\sqrt{-1}$$
$$y - \frac{2}{3} = \pm i$$
* Add $\frac{2}{3}$ to both sides:
$$y = \frac{2}{3} \pm i$$

Final Answer:
1) $x = \frac{7}{3}, x = -7$
2) $p = \frac{3}{5}, p = 2$
3) $l = 8, l = -1$
4) $t = -5, t = -3$
5) $s = \frac{-3 \pm 4\sqrt{19}}{5}$
6) $r = 2 \pm \sqrt{102}$
7) $v = \frac{3 \pm \sqrt{111}}{3}$
8) $y = \frac{2}{3} \pm i$