0=2x^2-18x+9

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



0=2x^2-18x+9
We move all terms to the left:
0-(2x^2-18x+9)=0
We add all the numbers together, and all the variables
-(2x^2-18x+9)=0
We get rid of parentheses
-2x^2+18x-9=0
a = -2; b = 18; c = -9;
Δ = b2-4ac
Δ = 182-4·(-2)·(-9)
Δ = 252
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:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{7}}{2*-2}=\frac{-18-6\sqrt{7}}{-4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{7}}{2*-2}=\frac{-18+6\sqrt{7}}{-4} $

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