2x^2+22x+28=-10x

Solution for 2x^2+22x+28=-10x equation:

2x^2+22x+28=-10x

We move all terms to the left:

2x^2+22x+28-(-10x)=0

We get rid of parentheses

2x^2+22x+10x+28=0

We add all the numbers together, and all the variables

2x^2+32x+28=0

a = 2; b = 32; c = +28;
Δ = b2-4ac
Δ = 322-4·2·28
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-20\sqrt{2}}{2*2}=\frac{-32-20\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+20\sqrt{2}}{2*2}=\frac{-32+20\sqrt{2}}{4}
$