a^2-4a=15

Solution for a^2-4a=15 equation:

a^2-4a=15

We move all terms to the left:

a^2-4a-(15)=0

a = 1; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·1·(-15)
Δ = 76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{19}}{2*1}=\frac{4-2\sqrt{19}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{19}}{2*1}=\frac{4+2\sqrt{19}}{2}
$