0=2x^2+21x-198

Solution for 0=2x^2+21x-198 equation:

0=2x^2+21x-198

We move all terms to the left:

0-(2x^2+21x-198)=0

We add all the numbers together, and all the variables

-(2x^2+21x-198)=0

We get rid of parentheses

-2x^2-21x+198=0

a = -2; b = -21; c = +198;
Δ = b2-4ac
Δ = -212-4·(-2)·198
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-45}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+45}{2*-2}=\frac{66}{-4}
=-16+1/2
$