a^2+16a=225

Solution for a^2+16a=225 equation:

a^2+16a=225

We move all terms to the left:

a^2+16a-(225)=0

a = 1; b = 16; c = -225;
Δ = b2-4ac
Δ = 162-4·1·(-225)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-34}{2*1}=\frac{-50}{2}
=-25
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+34}{2*1}=\frac{18}{2}
=9
$