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10q^2+23q=0
a = 10; b = 23; c = 0;
Δ = b2-4ac
Δ = 232-4·10·0
Δ = 529
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{529}=23$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-23}{2*10}=\frac{-46}{20} =-2+3/10 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+23}{2*10}=\frac{0}{20} =0 $
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