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3t^2+37t-14=0
a = 3; b = 37; c = -14;
Δ = b2-4ac
Δ = 372-4·3·(-14)
Δ = 1537
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{-(37)-\sqrt{1537}}{2*3}=\frac{-37-\sqrt{1537}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(37)+\sqrt{1537}}{2*3}=\frac{-37+\sqrt{1537}}{6} $
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