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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

40x^2+440x-160=0

a = 40; b = 440; c = -160;
Δ = b2-4ac
Δ = 4402-4·40·(-160)
Δ = 219200
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:

$\sqrt{\Delta}=\sqrt{219200}=\sqrt{1600*137}=\sqrt{1600}*\sqrt{137}=40\sqrt{137}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(440)-40\sqrt{137}}{2*40}=\frac{-440-40\sqrt{137}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(440)+40\sqrt{137}}{2*40}=\frac{-440+40\sqrt{137}}{80}
$