# How do you use the quadratic formula to solve 3x^2- 8x -15 =0 ?

Feb 12, 2017

$x = 3.94 \text{ and } - 1.27$ to 2 decimal places (approx. solution)

#### Explanation:

This is really worth trying to remember. I did it many years ago by making a point of writing it out for every question I attempted. Repetition is the key.

The standard form is:

$y = a {x}^{2} + b x + c \text{ }$ where $\text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
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Given: $3 {x}^{2} - 8 x - 15 = 0$

Set $a = 3 \text{; "b=-8"; } c = - 15$

So by substitution we have:

$x = \frac{+ 8 \pm \sqrt{{\left(- 8\right)}^{2} - 4 \left(3\right) \left(- 15\right)}}{2 \left(3\right)}$

$x = \frac{8 \pm \sqrt{64 + 180}}{6}$

$x = \frac{4}{3} \pm \frac{\sqrt{244}}{6}$

$x = \frac{4}{3} \pm \frac{\sqrt{{2}^{2} \times 61}}{6}$

$x = \frac{4}{3} \pm \frac{\sqrt{61}}{3} \text{ }$ This is an exact solution

$x = 3.94 \text{ and } - 1.27$ to 2 decimal places (approx. solution) 