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-12t^2+4t+40=0
a = -12; b = 4; c = +40;
Δ = b2-4ac
Δ = 42-4·(-12)·40
Δ = 1936
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{1936}=44$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-44}{2*-12}=\frac{-48}{-24} =+2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+44}{2*-12}=\frac{40}{-24} =-1+2/3 $
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