2x^2+16x-82=0

Solution for 2x^2+16x-82=0 equation:

2x^2+16x-82=0

a = 2; b = 16; c = -82;
Δ = b2-4ac
Δ = 162-4·2·(-82)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{57}}{2*2}=\frac{-16-4\sqrt{57}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{57}}{2*2}=\frac{-16+4\sqrt{57}}{4}
$