4x-4*5x+4=20x^2-4x-16

Solution for 4x-4*5x+4=20x^2-4x-16 equation:

4x-4*5x+4=20x^2-4x-16

We move all terms to the left:

4x-4*5x+4-(20x^2-4x-16)=0

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-20x^2-12x+20=0

a = -20; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·(-20)·20
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{109}}{2*-20}=\frac{12-4\sqrt{109}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{109}}{2*-20}=\frac{12+4\sqrt{109}}{-40}
$