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t^2+20t-96=0
a = 1; b = 20; c = -96;
Δ = b2-4ac
Δ = 202-4·1·(-96)
Δ = 784
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{784}=28$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-28}{2*1}=\frac{-48}{2} =-24 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+28}{2*1}=\frac{8}{2} =4 $
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