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42q^2=16q
We move all terms to the left:
42q^2-(16q)=0
a = 42; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·42·0
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$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*42}=\frac{0}{84} =0 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*42}=\frac{32}{84} =8/21 $
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