(x-2)2+10=-x2-8x

Solution for (x-2)2+10=-x2-8x equation:

(x-2)2+10=-x2-8x

We move all terms to the left:

(x-2)2+10-(-x2-8x)=0

We add all the numbers together, and all the variables

-(-1x^2-8x)+(x-2)2+10=0

We multiply parentheses

-(-1x^2-8x)+2x-4+10=0

We get rid of parentheses

1x^2+8x+2x-4+10=0

We add all the numbers together, and all the variables

x^2+10x+6=0

a = 1; b = 10; c = +6;
Δ = b2-4ac
Δ = 102-4·1·6
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{19}}{2*1}=\frac{-10-2\sqrt{19}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{19}}{2*1}=\frac{-10+2\sqrt{19}}{2}
$