5x^2+26x+8=0

Solution for 5x^2+26x+8=0 equation:

5x^2+26x+8=0

a = 5; b = 26; c = +8;
Δ = b2-4ac
Δ = 262-4·5·8
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{129}}{2*5}=\frac{-26-2\sqrt{129}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{129}}{2*5}=\frac{-26+2\sqrt{129}}{10}
$