3±√21. The equation [tex]y^{2}-6y-12=0[/tex] has two possible solutions 3+√21 y 3-√21.
If we have a general quadratic equation [tex]ay^{2} +by+c=0[/tex] we can solves the equation by completing the square. First, we divide the quadratic equation by a, we obtain [tex]y^{2} +\frac{b}{a} y+\frac{c}{a} =0[/tex].
For this problem, we have [tex]y^{2}-6y-12=0[/tex]
We can skipped division in this example since the coefficient of [tex]x^{2}[/tex] is 1.
Move the term c to the right side of the equation
[tex]y^{2}-6y=12[/tex]
Completing the square on the left side of the equation and balance this by adding the same number to the right side of the equation, with b = -6.
[tex](\frac{b}{2})^{2} =(\frac{-6}{2})^{2}=(-3)^{2} =9[/tex]
[tex]y^{2}-6y+9=12+9[/tex]
[tex](y-3)^{2}=21[/tex]
Take the square root on both sides of the equation:
y - 3 = ±√21
Add 3 from both sides:
y = 3 ± √21
