# How do you solve the quadratic x^2+3x-28=0 using any method?

Aug 31, 2016

$x = 4 , - 7$

#### Explanation:

${x}^{2} + 3 x - 28 = 0$

${x}^{2} + 7 x - 4 x - 28 = 0$

$x \left(x + 7\right) - 4 \left(x + 7\right) = 0$

$\left(x + 7\right) \left(x - 4\right) = 0$

Either $\left(x + 7\right) = 0 , \mathmr{and} \left(x - 4\right) = 0$

If $x + 7 = 0$

$x = - 7$

If $x - 7 = 0$

$x = 4$

$x = 4 , - 7$