2x^2+18x+96=-10x

Solution for 2x^2+18x+96=-10x equation:

2x^2+18x+96=-10x

We move all terms to the left:

2x^2+18x+96-(-10x)=0

We get rid of parentheses

2x^2+18x+10x+96=0

We add all the numbers together, and all the variables

2x^2+28x+96=0

a = 2; b = 28; c = +96;
Δ = b2-4ac
Δ = 282-4·2·96
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*2}=\frac{-24}{4}
=-6
$