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y^2=2y+5
We move all terms to the left:
y^2-(2y+5)=0
We get rid of parentheses
y^2-2y-5=0
a = 1; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·1·(-5)
Δ = 24
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{6}}{2*1}=\frac{2-2\sqrt{6}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{6}}{2*1}=\frac{2+2\sqrt{6}}{2} $
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