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22q^2+35q=0
a = 22; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·22·0
Δ = 1225
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{1225}=35$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*22}=\frac{-70}{44} =-1+13/22 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*22}=\frac{0}{44} =0 $
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