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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = -4; c = +10;
Δ = b2-4ac
Δ = -42-4·(-1)·10
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{14}}{2*-1}=\frac{4-2\sqrt{14}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{14}}{2*-1}=\frac{4+2\sqrt{14}}{-2}
$