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x^2=-18x+18
We move all terms to the left:
x^2-(-18x+18)=0
We get rid of parentheses
x^2+18x-18=0
a = 1; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·1·(-18)
Δ = 396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{11}}{2*1}=\frac{-18-6\sqrt{11}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{11}}{2*1}=\frac{-18+6\sqrt{11}}{2} $
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