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5t^2+39t+54=0
a = 5; b = 39; c = +54;
Δ = b2-4ac
Δ = 392-4·5·54
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-21}{2*5}=\frac{-60}{10} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+21}{2*5}=\frac{-18}{10} =-1+4/5 $
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