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3q^2+8q-16=0
a = 3; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·3·(-16)
Δ = 256
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{256}=16$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*3}=\frac{-24}{6} =-4 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*3}=\frac{8}{6} =1+1/3 $
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