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3y^{2}=9
Add 9 to both sides. Anything plus zero gives itself.
y^{2}=\frac{9}{3}
Divide both sides by 3.
y^{2}=3
Divide 9 by 3 to get 3.
y=\sqrt{3} y=-\sqrt{3}
Take the square root of both sides of the equation.
3y^{2}-9=0
Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.
y=\frac{0±\sqrt{0^{2}-4\times 3\left(-9\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{0±\sqrt{-4\times 3\left(-9\right)}}{2\times 3}
Square 0.
y=\frac{0±\sqrt{-12\left(-9\right)}}{2\times 3}
Multiply -4 times 3.
y=\frac{0±\sqrt{108}}{2\times 3}
Multiply -12 times -9.
y=\frac{0±6\sqrt{3}}{2\times 3}
Take the square root of 108.
y=\frac{0±6\sqrt{3}}{6}
Multiply 2 times 3.
y=\sqrt{3}
Now solve the equation y=\frac{0±6\sqrt{3}}{6} when ± is plus.
y=-\sqrt{3}
Now solve the equation y=\frac{0±6\sqrt{3}}{6} when ± is minus.
y=\sqrt{3} y=-\sqrt{3}
The equation is now solved.