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3y^2-12y+9=1
We move all terms to the left:
3y^2-12y+9-(1)=0
We add all the numbers together, and all the variables
3y^2-12y+8=0
a = 3; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·3·8
Δ = 48
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*3}=\frac{12-4\sqrt{3}}{6} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*3}=\frac{12+4\sqrt{3}}{6} $
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