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29-12y+y^2=0
a = 1; b = -12; c = +29;
Δ = b2-4ac
Δ = -122-4·1·29
Δ = 28
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{7}}{2*1}=\frac{12-2\sqrt{7}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{7}}{2*1}=\frac{12+2\sqrt{7}}{2} $
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