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10t^2+19t=0
a = 10; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·10·0
Δ = 361
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{361}=19$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*10}=\frac{-38}{20} =-1+9/10 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*10}=\frac{0}{20} =0 $
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