2x^2+36x-18=0

Solution for 2x^2+36x-18=0 equation:

2x^2+36x-18=0

a = 2; b = 36; c = -18;
Δ = b2-4ac
Δ = 362-4·2·(-18)
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12\sqrt{10}}{2*2}=\frac{-36-12\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12\sqrt{10}}{2*2}=\frac{-36+12\sqrt{10}}{4}
$