4c^2+28c=120

Solution for 4c^2+28c=120 equation:

4c^2+28c=120

We move all terms to the left:

4c^2+28c-(120)=0

a = 4; b = 28; c = -120;
Δ = b2-4ac
Δ = 282-4·4·(-120)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-52}{2*4}=\frac{-80}{8}
=-10
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+52}{2*4}=\frac{24}{8}
=3
$