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4t^2-19t-5=0
a = 4; b = -19; c = -5;
Δ = b2-4ac
Δ = -192-4·4·(-5)
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*4}=\frac{-2}{8} =-1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*4}=\frac{40}{8} =5 $
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