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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x^2+19x+42=0

a = 2; b = 19; c = +42;
Δ = b2-4ac
Δ = 192-4·2·42
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*2}=\frac{-14}{4}
=-3+1/2
$