2x^2+27x-160=0

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

2x^2+27x-160=0

a = 2; b = 27; c = -160;
Δ = b2-4ac
Δ = 272-4·2·(-160)
Δ = 2009
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:

$\sqrt{\Delta}=\sqrt{2009}=\sqrt{49*41}=\sqrt{49}*\sqrt{41}=7\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-7\sqrt{41}}{2*2}=\frac{-27-7\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+7\sqrt{41}}{2*2}=\frac{-27+7\sqrt{41}}{4}
$