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

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

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

We add all the numbers together, and all the variables

2x^2-9x=0

a = 2; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·2·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*2}=\frac{18}{4}
=4+1/2
$