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9x^2-18x=36
We move all terms to the left:
9x^2-18x-(36)=0
a = 9; b = -18; c = -36;
Δ = b2-4ac
Δ = -182-4·9·(-36)
Δ = 1620
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18\sqrt{5}}{2*9}=\frac{18-18\sqrt{5}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18\sqrt{5}}{2*9}=\frac{18+18\sqrt{5}}{18} $
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