4m^2+4m=120

Solution for 4m^2+4m=120 equation:

4m^2+4m=120

We move all terms to the left:

4m^2+4m-(120)=0

a = 4; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·4·(-120)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*4}=\frac{-48}{8}
=-6
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*4}=\frac{40}{8}
=5
$