x^2+(2x=5-5)^2=9

Solution for x^2+(2x=5-5)^2=9 equation:

x^2+(2x=5-5)^2=9

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2+2x-0^2=0

We add all the numbers together, and all the variables

x^2+2x=0

a = 1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*1}=\frac{0}{2}
=0
$