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8q^2+9q=0
a = 8; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·8·0
Δ = 81
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{81}=9$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*8}=\frac{-18}{16} =-1+1/8 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*8}=\frac{0}{16} =0 $
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