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12x^2+24x-96=0
a = 12; b = 24; c = -96;
Δ = b2-4ac
Δ = 242-4·12·(-96)
Δ = 5184
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{5184}=72$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-72}{2*12}=\frac{-96}{24} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+72}{2*12}=\frac{48}{24} =2 $
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