9x^2-160x+50=0

Solution for 9x^2-160x+50=0 equation:

9x^2-160x+50=0

a = 9; b = -160; c = +50;
Δ = b2-4ac
Δ = -1602-4·9·50
Δ = 23800
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:

$\sqrt{\Delta}=\sqrt{23800}=\sqrt{100*238}=\sqrt{100}*\sqrt{238}=10\sqrt{238}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-10\sqrt{238}}{2*9}=\frac{160-10\sqrt{238}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+10\sqrt{238}}{2*9}=\frac{160+10\sqrt{238}}{18}
$