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y2-6y+1=0
We add all the numbers together, and all the variables
y^2-6y+1=0
a = 1; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·1·1
Δ = 32
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{2}}{2*1}=\frac{6-4\sqrt{2}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{2}}{2*1}=\frac{6+4\sqrt{2}}{2} $
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