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4q^2=-8q
We move all terms to the left:
4q^2-(-8q)=0
We get rid of parentheses
4q^2+8q=0
a = 4; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·4·0
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$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*4}=\frac{-16}{8} =-2 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*4}=\frac{0}{8} =0 $
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