9x^2+26x+5=0

Solution for 9x^2+26x+5=0 equation:

9x^2+26x+5=0

a = 9; b = 26; c = +5;
Δ = b2-4ac
Δ = 262-4·9·5
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-4\sqrt{31}}{2*9}=\frac{-26-4\sqrt{31}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+4\sqrt{31}}{2*9}=\frac{-26+4\sqrt{31}}{18}
$