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q^2=63
We move all terms to the left:
q^2-(63)=0
a = 1; b = 0; c = -63;
Δ = b2-4ac
Δ = 02-4·1·(-63)
Δ = 252
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}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*1}=\frac{0-6\sqrt{7}}{2} =-\frac{6\sqrt{7}}{2} =-3\sqrt{7} $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*1}=\frac{0+6\sqrt{7}}{2} =\frac{6\sqrt{7}}{2} =3\sqrt{7} $
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