4x^2-12x+16=36

Solution for 4x^2-12x+16=36 equation:

4x^2-12x+16=36

We move all terms to the left:

4x^2-12x+16-(36)=0

We add all the numbers together, and all the variables

4x^2-12x-20=0

a = 4; b = -12; c = -20;
Δ = b2-4ac
Δ = -122-4·4·(-20)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{29}}{2*4}=\frac{12-4\sqrt{29}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{29}}{2*4}=\frac{12+4\sqrt{29}}{8}
$