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q^2=-42q
We move all terms to the left:
q^2-(-42q)=0
We get rid of parentheses
q^2+42q=0
a = 1; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·1·0
Δ = 1764
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{1764}=42$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*1}=\frac{-84}{2} =-42 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*1}=\frac{0}{2} =0 $
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