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14t^2+33t=0
a = 14; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·14·0
Δ = 1089
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{1089}=33$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*14}=\frac{-66}{28} =-2+5/14 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*14}=\frac{0}{28} =0 $
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