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3x^2-36x=96=0
We move all terms to the left:
3x^2-36x-(96)=0
a = 3; b = -36; c = -96;
Δ = b2-4ac
Δ = -362-4·3·(-96)
Δ = 2448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2448}=\sqrt{144*17}=\sqrt{144}*\sqrt{17}=12\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12\sqrt{17}}{2*3}=\frac{36-12\sqrt{17}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12\sqrt{17}}{2*3}=\frac{36+12\sqrt{17}}{6} $
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