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6x^2-4x=18
We move all terms to the left:
6x^2-4x-(18)=0
a = 6; b = -4; c = -18;
Δ = b2-4ac
Δ = -42-4·6·(-18)
Δ = 448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{7}}{2*6}=\frac{4-8\sqrt{7}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{7}}{2*6}=\frac{4+8\sqrt{7}}{12} $
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