2x^2-40x+200=02

Solution for 2x^2-40x+200=02 equation:

2x^2-40x+200=02

We move all terms to the left:

2x^2-40x+200-(02)=0

We add all the numbers together, and all the variables

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
$