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2t^2+27t-120=0
a = 2; b = 27; c = -120;
Δ = b2-4ac
Δ = 272-4·2·(-120)
Δ = 1689
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{-(27)-\sqrt{1689}}{2*2}=\frac{-27-\sqrt{1689}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{1689}}{2*2}=\frac{-27+\sqrt{1689}}{4} $
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