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12t^2-12=0
a = 12; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·12·(-12)
Δ = 576
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{576}=24$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*12}=\frac{-24}{24} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*12}=\frac{24}{24} =1 $
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