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3k^2+12k-96=0
a = 3; b = 12; c = -96;
Δ = b2-4ac
Δ = 122-4·3·(-96)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1296}=36$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36}{2*3}=\frac{-48}{6} =-8 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36}{2*3}=\frac{24}{6} =4 $
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