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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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-12}{2*3}=\frac{-48}{6} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+12}{2*3}=\frac{-24}{6} =-4 $
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