-4n(2n-5)=-28

Solution for -4n(2n-5)=-28 equation:

-4n(2n-5)=-28

We move all terms to the left:

-4n(2n-5)-(-28)=0

We add all the numbers together, and all the variables

-4n(2n-5)+28=0

We multiply parentheses

-8n^2+20n+28=0

a = -8; b = 20; c = +28;
Δ = b2-4ac
Δ = 202-4·(-8)·28
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*-8}=\frac{-56}{-16}
=3+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+36}{2*-8}=\frac{16}{-16}
=-1
$