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5t^2+4t-28=0
a = 5; b = 4; c = -28;
Δ = b2-4ac
Δ = 42-4·5·(-28)
Δ = 576
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{576}=24$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*5}=\frac{-28}{10} =-2+4/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*5}=\frac{20}{10} =2 $
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