20x=4x^2+24

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

20x=4x^2+24

We move all terms to the left:

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

We get rid of parentheses

-4x^2+20x-24=0

a = -4; b = 20; c = -24;
Δ = b2-4ac
Δ = 202-4·(-4)·(-24)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*-4}=\frac{-24}{-8}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*-4}=\frac{-16}{-8}
=+2
$