5x^2+2x-256=0

Solution for 5x^2+2x-256=0 equation:

5x^2+2x-256=0

a = 5; b = 2; c = -256;
Δ = b2-4ac
Δ = 22-4·5·(-256)
Δ = 5124
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:

$\sqrt{\Delta}=\sqrt{5124}=\sqrt{4*1281}=\sqrt{4}*\sqrt{1281}=2\sqrt{1281}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1281}}{2*5}=\frac{-2-2\sqrt{1281}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1281}}{2*5}=\frac{-2+2\sqrt{1281}}{10}
$