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