-2x^2+13x+32=0

Solution for -2x^2+13x+32=0 equation:

-2x^2+13x+32=0

a = -2; b = 13; c = +32;
Δ = b2-4ac
Δ = 132-4·(-2)·32
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5\sqrt{17}}{2*-2}=\frac{-13-5\sqrt{17}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5\sqrt{17}}{2*-2}=\frac{-13+5\sqrt{17}}{-4}
$