2x+2=2x^2-4x+5

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

2x+2=2x^2-4x+5

We move all terms to the left:

2x+2-(2x^2-4x+5)=0

We get rid of parentheses

-2x^2+2x+4x-5+2=0

We add all the numbers together, and all the variables

-2x^2+6x-3=0

a = -2; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·(-2)·(-3)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{3}}{2*-2}=\frac{-6-2\sqrt{3}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{3}}{2*-2}=\frac{-6+2\sqrt{3}}{-4}
$