x^2+15x-252=0

Solution for x^2+15x-252=0 equation:

x^2+15x-252=0

a = 1; b = 15; c = -252;
Δ = b2-4ac
Δ = 152-4·1·(-252)
Δ = 1233
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1233}=\sqrt{9*137}=\sqrt{9}*\sqrt{137}=3\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{137}}{2*1}=\frac{-15-3\sqrt{137}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{137}}{2*1}=\frac{-15+3\sqrt{137}}{2}
$