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35t^2+11t=0
a = 35; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·35·0
Δ = 121
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{121}=11$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*35}=\frac{-22}{70} =-11/35 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*35}=\frac{0}{70} =0 $
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