10x^2+24x-128=0

Solution for 10x^2+24x-128=0 equation:

10x^2+24x-128=0

a = 10; b = 24; c = -128;
Δ = b2-4ac
Δ = 242-4·10·(-128)
Δ = 5696
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:

$\sqrt{\Delta}=\sqrt{5696}=\sqrt{64*89}=\sqrt{64}*\sqrt{89}=8\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{89}}{2*10}=\frac{-24-8\sqrt{89}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{89}}{2*10}=\frac{-24+8\sqrt{89}}{20}
$