0=2x^2+20x+18+

Solution for 0=2x^2+20x+18+ equation:

0=2x^2+20x+18+

We move all terms to the left:

0-(2x^2+20x+18+)=0

We add all the numbers together, and all the variables

-(2x^2+20x+18+)=0

We get rid of parentheses

-2x^2-20x-18-=0

We add all the numbers together, and all the variables

-2x^2-20x=0

a = -2; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-2)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-2}=\frac{40}{-4}
=-10
$