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5t^2-10t-40=0
a = 5; b = -10; c = -40;
Δ = b2-4ac
Δ = -102-4·5·(-40)
Δ = 900
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{900}=30$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-30}{2*5}=\frac{-20}{10} =-2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+30}{2*5}=\frac{40}{10} =4 $
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