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q2=28q=14
We move all terms to the left:
q2-(28q)=0
We add all the numbers together, and all the variables
q^2-28q=0
a = 1; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·1·0
Δ = 784
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{784}=28$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*1}=\frac{0}{2} =0 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*1}=\frac{56}{2} =28 $
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