m^2+4m=15

Solution for m^2+4m=15 equation:

m^2+4m=15

We move all terms to the left:

m^2+4m-(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:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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