2x^2+18x=-40

Solution for 2x^2+18x=-40 equation:

2x^2+18x=-40

We move all terms to the left:

2x^2+18x-(-40)=0

We add all the numbers together, and all the variables

2x^2+18x+40=0

a = 2; b = 18; c = +40;
Δ = b2-4ac
Δ = 182-4·2·40
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2}{2*2}=\frac{-20}{4}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2}{2*2}=\frac{-16}{4}
=-4
$