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y^2-2y-5=3
We move all terms to the left:
y^2-2y-5-(3)=0
We add all the numbers together, and all the variables
y^2-2y-8=0
a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2} =-2 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2} =4 $
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