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4t^2+11t-20=0
a = 4; b = 11; c = -20;
Δ = b2-4ac
Δ = 112-4·4·(-20)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*4}=\frac{-32}{8} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*4}=\frac{10}{8} =1+1/4 $
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