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2t^2+63t+145=0
a = 2; b = 63; c = +145;
Δ = b2-4ac
Δ = 632-4·2·145
Δ = 2809
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{2809}=53$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-53}{2*2}=\frac{-116}{4} =-29 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+53}{2*2}=\frac{-10}{4} =-2+1/2 $
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