16=4x^2-20x+24

Solution for 16=4x^2-20x+24 equation:

16=4x^2-20x+24

We move all terms to the left:

16-(4x^2-20x+24)=0

We get rid of parentheses

-4x^2+20x-24+16=0

We add all the numbers together, and all the variables

-4x^2+20x-8=0

a = -4; b = 20; c = -8;
Δ = b2-4ac
Δ = 202-4·(-4)·(-8)
Δ = 272
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{17}}{2*-4}=\frac{-20-4\sqrt{17}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{17}}{2*-4}=\frac{-20+4\sqrt{17}}{-8}
$