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3x^2-96x=0
a = 3; b = -96; c = 0;
Δ = b2-4ac
Δ = -962-4·3·0
Δ = 9216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9216}=96$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-96}{2*3}=\frac{0}{6} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+96}{2*3}=\frac{192}{6} =32 $
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