-20x^2+22x+160=0

Solution for -20x^2+22x+160=0 equation:

-20x^2+22x+160=0

a = -20; b = 22; c = +160;
Δ = b2-4ac
Δ = 222-4·(-20)·160
Δ = 13284
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:

$\sqrt{\Delta}=\sqrt{13284}=\sqrt{324*41}=\sqrt{324}*\sqrt{41}=18\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18\sqrt{41}}{2*-20}=\frac{-22-18\sqrt{41}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18\sqrt{41}}{2*-20}=\frac{-22+18\sqrt{41}}{-40}
$