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6q^2-25q+4=0
a = 6; b = -25; c = +4;
Δ = b2-4ac
Δ = -252-4·6·4
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$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-23}{2*6}=\frac{2}{12} =1/6 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+23}{2*6}=\frac{48}{12} =4 $
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