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21t^2+13t+2=0
a = 21; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·21·2
Δ = 1
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{1}=1$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*21}=\frac{-14}{42} =-1/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*21}=\frac{-12}{42} =-2/7 $
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