8x^2+16x-92=0

Solution for 8x^2+16x-92=0 equation:

8x^2+16x-92=0

a = 8; b = 16; c = -92;
Δ = b2-4ac
Δ = 162-4·8·(-92)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-40\sqrt{2}}{2*8}=\frac{-16-40\sqrt{2}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+40\sqrt{2}}{2*8}=\frac{-16+40\sqrt{2}}{16}
$