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6t^2+17t+5=0
a = 6; b = 17; c = +5;
Δ = b2-4ac
Δ = 172-4·6·5
Δ = 169
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{169}=13$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*6}=\frac{-30}{12} =-2+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*6}=\frac{-4}{12} =-1/3 $
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