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y2=6y+4
We move all terms to the left:
y2-(6y+4)=0
We add all the numbers together, and all the variables
y^2-(6y+4)=0
We get rid of parentheses
y^2-6y-4=0
a = 1; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·1·(-4)
Δ = 52
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*1}=\frac{6-2\sqrt{13}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*1}=\frac{6+2\sqrt{13}}{2} $
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