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2+y2=5
We move all terms to the left:
2+y2-(5)=0
We add all the numbers together, and all the variables
y^2-3=0
a = 1; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·1·(-3)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*1}=\frac{0-2\sqrt{3}}{2} =-\frac{2\sqrt{3}}{2} =-\sqrt{3} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*1}=\frac{0+2\sqrt{3}}{2} =\frac{2\sqrt{3}}{2} =\sqrt{3} $
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