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11y^{2}+y=2
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
11y^{2}+y-2=2-2
Subtract 2 from both sides of the equation.
11y^{2}+y-2=0
Subtracting 2 from itself leaves 0.
y=\frac{-1±\sqrt{1^{2}-4\times 11\left(-2\right)}}{2\times 11}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, 1 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-1±\sqrt{1-4\times 11\left(-2\right)}}{2\times 11}
Square 1.
y=\frac{-1±\sqrt{1-44\left(-2\right)}}{2\times 11}
Multiply -4 times 11.
y=\frac{-1±\sqrt{1+88}}{2\times 11}
Multiply -44 times -2.
y=\frac{-1±\sqrt{89}}{2\times 11}
Add 1 to 88.
y=\frac{-1±\sqrt{89}}{22}
Multiply 2 times 11.
y=\frac{\sqrt{89}-1}{22}
Now solve the equation y=\frac{-1±\sqrt{89}}{22} when ± is plus. Add -1 to \sqrt{89}.
y=\frac{-\sqrt{89}-1}{22}
Now solve the equation y=\frac{-1±\sqrt{89}}{22} when ± is minus. Subtract \sqrt{89} from -1.
y=\frac{\sqrt{89}-1}{22} y=\frac{-\sqrt{89}-1}{22}
The equation is now solved.
11y^{2}+y=2
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
\frac{11y^{2}+y}{11}=\frac{2}{11}
Divide both sides by 11.
y^{2}+\frac{1}{11}y=\frac{2}{11}
Dividing by 11 undoes the multiplication by 11.
y^{2}+\frac{1}{11}y+\left(\frac{1}{22}\right)^{2}=\frac{2}{11}+\left(\frac{1}{22}\right)^{2}
Divide \frac{1}{11}, the coefficient of the x term, by 2 to get \frac{1}{22}. Then add the square of \frac{1}{22} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+\frac{1}{11}y+\frac{1}{484}=\frac{2}{11}+\frac{1}{484}
Square \frac{1}{22} by squaring both the numerator and the denominator of the fraction.
y^{2}+\frac{1}{11}y+\frac{1}{484}=\frac{89}{484}
Add \frac{2}{11} to \frac{1}{484} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y+\frac{1}{22}\right)^{2}=\frac{89}{484}
Factor y^{2}+\frac{1}{11}y+\frac{1}{484}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(y+\frac{1}{22}\right)^{2}}=\sqrt{\frac{89}{484}}
Take the square root of both sides of the equation.
y+\frac{1}{22}=\frac{\sqrt{89}}{22} y+\frac{1}{22}=-\frac{\sqrt{89}}{22}
Simplify.
y=\frac{\sqrt{89}-1}{22} y=\frac{-\sqrt{89}-1}{22}
Subtract \frac{1}{22} from both sides of the equation.