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13t^2+22t=0
a = 13; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·13·0
Δ = 484
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*13}=\frac{-44}{26} =-1+9/13 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*13}=\frac{0}{26} =0 $
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