8x^2+24x-64=0

Solution for 8x^2+24x-64=0 equation:

8x^2+24x-64=0

a = 8; b = 24; c = -64;
Δ = b2-4ac
Δ = 242-4·8·(-64)
Δ = 2624
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:

$\sqrt{\Delta}=\sqrt{2624}=\sqrt{64*41}=\sqrt{64}*\sqrt{41}=8\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{41}}{2*8}=\frac{-24-8\sqrt{41}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{41}}{2*8}=\frac{-24+8\sqrt{41}}{16}
$