160=40x^2+240x+200

Solution for 160=40x^2+240x+200 equation:

160=40x^2+240x+200

We move all terms to the left:

160-(40x^2+240x+200)=0

We get rid of parentheses

-40x^2-240x-200+160=0

We add all the numbers together, and all the variables

-40x^2-240x-40=0

a = -40; b = -240; c = -40;
Δ = b2-4ac
Δ = -2402-4·(-40)·(-40)
Δ = 51200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{51200}=\sqrt{25600*2}=\sqrt{25600}*\sqrt{2}=160\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-160\sqrt{2}}{2*-40}=\frac{240-160\sqrt{2}}{-80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+160\sqrt{2}}{2*-40}=\frac{240+160\sqrt{2}}{-80}
$