Solve 11y=3y^2-4

$$11 y = 3 y ^ { 2 } - 4$$

Solve for y

$$y=4$$

$$y=-\frac{1}{3}\approx -0.333333333$$

Steps Using Factoring By Grouping

Subtract 3y^{2} from both sides.

$$11y-3y^{2}=-4$$

Add 4 to both sides.

$$11y-3y^{2}+4=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3y^{2}+11y+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3y^{2}+ay+by+4. To find a and b, set up a system to be solved.

$$a+b=11 ab=-3\times 4=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 11.

$$a=12 b=-1$$

Rewrite -3y^{2}+11y+4 as \left(-3y^{2}+12y\right)+\left(-y+4\right).

$$\left(-3y^{2}+12y\right)+\left(-y+4\right)$$

Factor out 3y in -3y^{2}+12y.

$$3y\left(-y+4\right)-y+4$$

Factor out common term -y+4 by using distributive property.

$$\left(-y+4\right)\left(3y+1\right)$$

To find equation solutions, solve -y+4=0 and 3y+1=0.

$$y=4 y=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 3y^{2} from both sides.

$$11y-3y^{2}=-4$$

Add 4 to both sides.

$$11y-3y^{2}+4=0$$

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.

$$-3y^{2}+11y+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 11 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-11±\sqrt{11^{2}-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Square 11.

$$y=\frac{-11±\sqrt{121-4\left(-3\right)\times 4}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$y=\frac{-11±\sqrt{121+12\times 4}}{2\left(-3\right)}$$

Multiply 12 times 4.

$$y=\frac{-11±\sqrt{121+48}}{2\left(-3\right)}$$

Add 121 to 48.

$$y=\frac{-11±\sqrt{169}}{2\left(-3\right)}$$

Take the square root of 169.

$$y=\frac{-11±13}{2\left(-3\right)}$$

Multiply 2 times -3.

$$y=\frac{-11±13}{-6}$$

Now solve the equation y=\frac{-11±13}{-6} when ± is plus. Add -11 to 13.

$$y=\frac{2}{-6}$$

Reduce the fraction \frac{2}{-6} to lowest terms by extracting and canceling out 2.

$$y=-\frac{1}{3}$$

Now solve the equation y=\frac{-11±13}{-6} when ± is minus. Subtract 13 from -11.

$$y=-\frac{24}{-6}$$

Divide -24 by -6.

$$y=4$$

The equation is now solved.

$$y=-\frac{1}{3} y=4$$

Steps for Completing the Square

Subtract 3y^{2} from both sides.

$$11y-3y^{2}=-4$$

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.

$$-3y^{2}+11y=-4$$

Divide both sides by -3.

$$\frac{-3y^{2}+11y}{-3}=-\frac{4}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y^{2}+\frac{11}{-3}y=-\frac{4}{-3}$$

Divide 11 by -3.

$$y^{2}-\frac{11}{3}y=-\frac{4}{-3}$$

Divide -4 by -3.

$$y^{2}-\frac{11}{3}y=\frac{4}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{11}{3}y+\left(-\frac{11}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{11}{3}y+\frac{121}{36}=\frac{4}{3}+\frac{121}{36}$$

Add \frac{4}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{11}{3}y+\frac{121}{36}=\frac{169}{36}$$

Factor y^{2}-\frac{11}{3}y+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{11}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$y-\frac{11}{6}=\frac{13}{6} y-\frac{11}{6}=-\frac{13}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$y=4 y=-\frac{1}{3}$$