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q2+8q=0

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Solution for q2+8q=0 equation:



q2+8q=0
We add all the numbers together, and all the variables
q^2+8q=0
a = 1; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·1·0
Δ = 64
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{64}=8
q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*1}=\frac{-16}{2} =-8
q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*1}=\frac{0}{2} =0

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