3x^2+26x-128=0

Solution for 3x^2+26x-128=0 equation:

3x^2+26x-128=0

a = 3; b = 26; c = -128;
Δ = b2-4ac
Δ = 262-4·3·(-128)
Δ = 2212
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:

$\sqrt{\Delta}=\sqrt{2212}=\sqrt{4*553}=\sqrt{4}*\sqrt{553}=2\sqrt{553}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{553}}{2*3}=\frac{-26-2\sqrt{553}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{553}}{2*3}=\frac{-26+2\sqrt{553}}{6}
$