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3x^2-36x=-96
We move all terms to the left:
3x^2-36x-(-96)=0
We add all the numbers together, and all the variables
3x^2-36x+96=0
a = 3; b = -36; c = +96;
Δ = b2-4ac
Δ = -362-4·3·96
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12}{2*3}=\frac{24}{6} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12}{2*3}=\frac{48}{6} =8 $
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