b^2+5=17

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Solution for b^2+5=17 equation:



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 = b2-4ac
Δ = 02-4·1·(-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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