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t^2+21t-65=0
a = 1; b = 21; c = -65;
Δ = b2-4ac
Δ = 212-4·1·(-65)
Δ = 701
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{701}}{2*1}=\frac{-21-\sqrt{701}}{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{701}}{2*1}=\frac{-21+\sqrt{701}}{2} $
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