-2x^2+12x+40=0

Solution for -2x^2+12x+40=0 equation:

-2x^2+12x+40=0

a = -2; b = 12; c = +40;
Δ = b2-4ac
Δ = 122-4·(-2)·40
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{29}}{2*-2}=\frac{-12-4\sqrt{29}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{29}}{2*-2}=\frac{-12+4\sqrt{29}}{-4}
$