m^2+12m=20

Solution for m^2+12m=20 equation:

m^2+12m=20

We move all terms to the left:

m^2+12m-(20)=0

a = 1; b = 12; c = -20;
Δ = b2-4ac
Δ = 122-4·1·(-20)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{14}}{2*1}=\frac{-12-4\sqrt{14}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{14}}{2*1}=\frac{-12+4\sqrt{14}}{2}
$