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8q(q-2)=0
We multiply parentheses
8q^2-16q=0
a = 8; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·8·0
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$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*8}=\frac{0}{16} =0 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*8}=\frac{32}{16} =2 $
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