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-3t^2+21t-25=0
a = -3; b = 21; c = -25;
Δ = b2-4ac
Δ = 212-4·(-3)·(-25)
Δ = 141
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{-(21)-\sqrt{141}}{2*-3}=\frac{-21-\sqrt{141}}{-6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{141}}{2*-3}=\frac{-21+\sqrt{141}}{-6} $
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