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-4t^2-21+19t=0
a = -4; b = 19; c = -21;
Δ = b2-4ac
Δ = 192-4·(-4)·(-21)
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*-4}=\frac{-24}{-8} =+3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*-4}=\frac{-14}{-8} =1+3/4 $
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