-4n(2n-5)=-8n+20

Solution for -4n(2n-5)=-8n+20 equation:

-4n(2n-5)=-8n+20

We move all terms to the left:

-4n(2n-5)-(-8n+20)=0

We multiply parentheses

-8n^2+20n-(-8n+20)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -8; b = 28; c = -20;
Δ = b2-4ac
Δ = 282-4·(-8)·(-20)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12}{2*-8}=\frac{-40}{-16}
=2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12}{2*-8}=\frac{-16}{-16}
=1
$