2x^2+(10-2x)=22

Solution for 2x^2+(10-2x)=22 equation:

2x^2+(10-2x)=22

We move all terms to the left:

2x^2+(10-2x)-(22)=0

We add all the numbers together, and all the variables

2x^2+(-2x+10)-22=0

We get rid of parentheses

2x^2-2x+10-22=0

We add all the numbers together, and all the variables

2x^2-2x-12=0

a = 2; b = -2; c = -12;
Δ = b2-4ac
Δ = -22-4·2·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*2}=\frac{12}{4}
=3
$