2x^2-20x+52=16

Solution for 2x^2-20x+52=16 equation:

2x^2-20x+52=16

We move all terms to the left:

2x^2-20x+52-(16)=0

We add all the numbers together, and all the variables

2x^2-20x+36=0

a = 2; b = -20; c = +36;
Δ = b2-4ac
Δ = -202-4·2·36
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{7}}{2*2}=\frac{20-4\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{7}}{2*2}=\frac{20+4\sqrt{7}}{4}
$