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b^2+5=17

We move all terms to the left:

b^2+5-(17)=0

We add all the numbers together, and all the variables

b^2-12=0

a = 1; b = 0; c = -12;

Δ = b^{2}-4ac

Δ = 0^{2}-4·1·(-12)

Δ = 48

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{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*1}=\frac{0-4\sqrt{3}}{2} =-\frac{4\sqrt{3}}{2} =-2\sqrt{3} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*1}=\frac{0+4\sqrt{3}}{2} =\frac{4\sqrt{3}}{2} =2\sqrt{3} $

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