# How do you solve 15x^2+7x-55 using the quadratic formula?

May 5, 2016

$x = \frac{- 7 \pm \sqrt{3349}}{30}$

#### Explanation:

Strictly speaking, I guess you would like to solve $15 {x}^{2} + 7 x - 55 = 0$ or in other words find the zeros of $15 {x}^{2} + 7 x - 55$. You do not "solve" a quadratic expression.

That having been said, the equation $15 {x}^{2} + 7 x - 55 = 0$ is of the form $a {x}^{2} + b x + c = 0$ with $a = 15$, $b = 7$ and $c = 55$.

This has roots (solutions) given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 7 \pm \sqrt{{\left(- 7\right)}^{2} - \left(4 \cdot 15 \cdot \left(- 55\right)\right)}}{2 \cdot 15}$

$= \frac{- 7 \pm \sqrt{49 + 3300}}{30}$

$= \frac{- 7 \pm \sqrt{3349}}{30}$

The square root $\sqrt{3349}$ does not simplify further since the prime factorisation of $3349$ is $17 \cdot 197$, which contains no square factors.