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3q^2+25q+8=0
a = 3; b = 25; c = +8;
Δ = b2-4ac
Δ = 252-4·3·8
Δ = 529
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{529}=23$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*3}=\frac{-48}{6} =-8 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*3}=\frac{-2}{6} =-1/3 $
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