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3t^2-15t-103=0
a = 3; b = -15; c = -103;
Δ = b2-4ac
Δ = -152-4·3·(-103)
Δ = 1461
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{-(-15)-\sqrt{1461}}{2*3}=\frac{15-\sqrt{1461}}{6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{1461}}{2*3}=\frac{15+\sqrt{1461}}{6} $
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