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4t^2+12t-40=0
a = 4; b = 12; c = -40;
Δ = b2-4ac
Δ = 122-4·4·(-40)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*4}=\frac{-40}{8} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*4}=\frac{16}{8} =2 $
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