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y2-4y=-2
We move all terms to the left:
y2-4y-(-2)=0
We add all the numbers together, and all the variables
y^2-4y+2=0
a = 1; b = -4; c = +2;
Δ = b2-4ac
Δ = -42-4·1·2
Δ = 8
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*1}=\frac{4-2\sqrt{2}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*1}=\frac{4+2\sqrt{2}}{2} $
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