2x^2+64x-2112=0

Solution for 2x^2+64x-2112=0 equation:

2x^2+64x-2112=0

a = 2; b = 64; c = -2112;
Δ = b2-4ac
Δ = 642-4·2·(-2112)
Δ = 20992
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:

$\sqrt{\Delta}=\sqrt{20992}=\sqrt{256*82}=\sqrt{256}*\sqrt{82}=16\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-16\sqrt{82}}{2*2}=\frac{-64-16\sqrt{82}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+16\sqrt{82}}{2*2}=\frac{-64+16\sqrt{82}}{4}
$