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