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3q^2+17q+14=0
a = 3; b = 17; c = +14;
Δ = b2-4ac
Δ = 172-4·3·14
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-11}{2*3}=\frac{-28}{6} =-4+2/3 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+11}{2*3}=\frac{-6}{6} =-1 $
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