27x^2+160x-256=0

Solution for 27x^2+160x-256=0 equation:

27x^2+160x-256=0

a = 27; b = 160; c = -256;
Δ = b2-4ac
Δ = 1602-4·27·(-256)
Δ = 53248
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:

$\sqrt{\Delta}=\sqrt{53248}=\sqrt{4096*13}=\sqrt{4096}*\sqrt{13}=64\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-64\sqrt{13}}{2*27}=\frac{-160-64\sqrt{13}}{54}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+64\sqrt{13}}{2*27}=\frac{-160+64\sqrt{13}}{54}
$