5x^2+18x-9=0.

Solution for 5x^2+18x-9=0. equation:

5x^2+18x-9=0.

a = 5; b = 18; c = -9;
Δ = b2-4ac
Δ = 182-4·5·(-9)
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{14}}{2*5}=\frac{-18-6\sqrt{14}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{14}}{2*5}=\frac{-18+6\sqrt{14}}{10}
$