=2x2+40x+198=0

Solution for =2x2+40x+198=0 equation:

=2x^2+40x+198=0

We move all terms to the left:

-(2x^2+40x+198)=0

We get rid of parentheses

-2x^2-40x-198=0

a = -2; b = -40; c = -198;
Δ = b2-4ac
Δ = -402-4·(-2)·(-198)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4}{2*-2}=\frac{36}{-4}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4}{2*-2}=\frac{44}{-4}
=-11
$