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b^2=39
We move all terms to the left:
b^2-(39)=0
a = 1; b = 0; c = -39;
Δ = b2-4ac
Δ = 02-4·1·(-39)
Δ = 156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{39}}{2*1}=\frac{0-2\sqrt{39}}{2} =-\frac{2\sqrt{39}}{2} =-\sqrt{39} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{39}}{2*1}=\frac{0+2\sqrt{39}}{2} =\frac{2\sqrt{39}}{2} =\sqrt{39} $
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