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3t^2+17t+10=0
a = 3; b = 17; c = +10;
Δ = b2-4ac
Δ = 172-4·3·10
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*3}=\frac{-30}{6} =-5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*3}=\frac{-4}{6} =-2/3 $
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