8x^2+32x-1=0

Solution for 8x^2+32x-1=0 equation:

8x^2+32x-1=0

a = 8; b = 32; c = -1;
Δ = b2-4ac
Δ = 322-4·8·(-1)
Δ = 1056
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:

$\sqrt{\Delta}=\sqrt{1056}=\sqrt{16*66}=\sqrt{16}*\sqrt{66}=4\sqrt{66}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{66}}{2*8}=\frac{-32-4\sqrt{66}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{66}}{2*8}=\frac{-32+4\sqrt{66}}{16}
$