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6q^2-4q-16=0
a = 6; b = -4; c = -16;
Δ = b2-4ac
Δ = -42-4·6·(-16)
Δ = 400
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{400}=20$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*6}=\frac{-16}{12} =-1+1/3 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*6}=\frac{24}{12} =2 $
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