2x^2+8=-10x

Solution for 2x^2+8=-10x equation:

2x^2+8=-10x

We move all terms to the left:

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

We get rid of parentheses

2x^2+10x+8=0

a = 2; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·2·8
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*2}=\frac{-4}{4}
=-1
$