2x^2=44+20x

Solution for 2x^2=44+20x equation:

2x^2=44+20x

We move all terms to the left:

2x^2-(44+20x)=0

We add all the numbers together, and all the variables

2x^2-(20x+44)=0

We get rid of parentheses

2x^2-20x-44=0

a = 2; b = -20; c = -44;
Δ = b2-4ac
Δ = -202-4·2·(-44)
Δ = 752
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:

$\sqrt{\Delta}=\sqrt{752}=\sqrt{16*47}=\sqrt{16}*\sqrt{47}=4\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{47}}{2*2}=\frac{20-4\sqrt{47}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{47}}{2*2}=\frac{20+4\sqrt{47}}{4}
$