16x^2+40x=160

Solution for 16x^2+40x=160 equation:

16x^2+40x=160

We move all terms to the left:

16x^2+40x-(160)=0

a = 16; b = 40; c = -160;
Δ = b2-4ac
Δ = 402-4·16·(-160)
Δ = 11840
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:

$\sqrt{\Delta}=\sqrt{11840}=\sqrt{64*185}=\sqrt{64}*\sqrt{185}=8\sqrt{185}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{185}}{2*16}=\frac{-40-8\sqrt{185}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{185}}{2*16}=\frac{-40+8\sqrt{185}}{32}
$