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3x^2-18x=135
We move all terms to the left:
3x^2-18x-(135)=0
a = 3; b = -18; c = -135;
Δ = b2-4ac
Δ = -182-4·3·(-135)
Δ = 1944
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18\sqrt{6}}{2*3}=\frac{18-18\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18\sqrt{6}}{2*3}=\frac{18+18\sqrt{6}}{6} $
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