4n^2+20n=24

Solution for 4n^2+20n=24 equation:

4n^2+20n=24

We move all terms to the left:

4n^2+20n-(24)=0

a = 4; b = 20; c = -24;
Δ = b2-4ac
Δ = 202-4·4·(-24)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*4}=\frac{-48}{8}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*4}=\frac{8}{8}
=1
$