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2x(x-3)=18
We move all terms to the left:
2x(x-3)-(18)=0
We multiply parentheses
2x^2-6x-18=0
a = 2; b = -6; c = -18;
Δ = b2-4ac
Δ = -62-4·2·(-18)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{5}}{2*2}=\frac{6-6\sqrt{5}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{5}}{2*2}=\frac{6+6\sqrt{5}}{4} $
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