2x^2+19x+40=-2x

Solution for 2x^2+19x+40=-2x equation:

2x^2+19x+40=-2x

We move all terms to the left:

2x^2+19x+40-(-2x)=0

We get rid of parentheses

2x^2+19x+2x+40=0

We add all the numbers together, and all the variables

2x^2+21x+40=0

a = 2; b = 21; c = +40;
Δ = b2-4ac
Δ = 212-4·2·40
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*2}=\frac{-32}{4}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*2}=\frac{-10}{4}
=-2+1/2
$