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-2.7t^2+40t-9.5=0
a = -2.7; b = 40; c = -9.5;
Δ = b2-4ac
Δ = 402-4·(-2.7)·(-9.5)
Δ = 1497.4
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-\sqrt{1497.4}}{2*-2.7}=\frac{-40-\sqrt{1497.4}}{-5.4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+\sqrt{1497.4}}{2*-2.7}=\frac{-40+\sqrt{1497.4}}{-5.4} $
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