198+198x=20x^2+40x

Solution for 198+198x=20x^2+40x equation:

198+198x=20x^2+40x

We move all terms to the left:

198+198x-(20x^2+40x)=0

We get rid of parentheses

-20x^2+198x-40x+198=0

We add all the numbers together, and all the variables

-20x^2+158x+198=0

a = -20; b = 158; c = +198;
Δ = b2-4ac
Δ = 1582-4·(-20)·198
Δ = 40804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{40804}=202$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(158)-202}{2*-20}=\frac{-360}{-40}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(158)+202}{2*-20}=\frac{44}{-40}
=-1+1/10
$