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9x^2-6=18x
We move all terms to the left:
9x^2-6-(18x)=0
a = 9; b = -18; c = -6;
Δ = b2-4ac
Δ = -182-4·9·(-6)
Δ = 540
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{15}}{2*9}=\frac{18-6\sqrt{15}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{15}}{2*9}=\frac{18+6\sqrt{15}}{18} $
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