w^2+15w-225=0

Solution for w^2+15w-225=0 equation:

w^2+15w-225=0

a = 1; b = 15; c = -225;
Δ = b2-4ac
Δ = 152-4·1·(-225)
Δ = 1125
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1125}=\sqrt{225*5}=\sqrt{225}*\sqrt{5}=15\sqrt{5}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15\sqrt{5}}{2*1}=\frac{-15-15\sqrt{5}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15\sqrt{5}}{2*1}=\frac{-15+15\sqrt{5}}{2}
$