4n^2=-20n+56

Solution for 4n^2=-20n+56 equation:

4n^2=-20n+56

We move all terms to the left:

4n^2-(-20n+56)=0

We get rid of parentheses

4n^2+20n-56=0

a = 4; b = 20; c = -56;
Δ = b2-4ac
Δ = 202-4·4·(-56)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-36}{2*4}=\frac{-56}{8}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+36}{2*4}=\frac{16}{8}
=2
$