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7t^2+2t-5=0
a = 7; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·7·(-5)
Δ = 144
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{144}=12$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-12}{2*7}=\frac{-14}{14} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*7}=\frac{10}{14} =5/7 $
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