5x^2+26x-23=0

Solution for 5x^2+26x-23=0 equation:

5x^2+26x-23=0

a = 5; b = 26; c = -23;
Δ = b2-4ac
Δ = 262-4·5·(-23)
Δ = 1136
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:

$\sqrt{\Delta}=\sqrt{1136}=\sqrt{16*71}=\sqrt{16}*\sqrt{71}=4\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4\sqrt{71}}{2*5}=\frac{-26-4\sqrt{71}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4\sqrt{71}}{2*5}=\frac{-26+4\sqrt{71}}{10}
$