# How do you write a quadratic equation in standard form that has two solutions, 6, and -10 with the leading coefficient one?

Sep 29, 2016

$y = {x}^{2} + 4 x - 60$

#### Explanation:

We want an equation which equals $0$ at the given points $6$ and $- 10$.

Our quadratic equation should be a product of expressions which are zero at the specified roots.
Consider $\left(x - 6\right) \cdot \left(x + 10\right) = 0$

This equality holds if $x = 6$ since
$\left(6 - 6\right) \cdot \left(6 + 10\right) = 0 \cdot 16 = 0$
And the equality holds if $x = - 10$ since
$\left(- 10 - 6\right) \cdot \left(- 10 + 10\right) = - 16 \cdot 0 = 0$

Expanding this equation by the FOIL method, we get:

${x}^{2} + 10 x - 6 x - 60$

Combining like terms, we find our solution:

${x}^{2} + 4 x - 60$