2x^2-18x-9=0

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

2x^2-18x-9=0

a = 2; b = -18; c = -9;
Δ = b2-4ac
Δ = -182-4·2·(-9)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{11}}{2*2}=\frac{18-6\sqrt{11}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{11}}{2*2}=\frac{18+6\sqrt{11}}{4}
$