Solve 11y-6=4y^2

$$11 y - 6 = 4 y ^ { 2 }$$

Solve for y

$$y=2$$

$$y=\frac{3}{4}=0.75$$

Steps Using Factoring By Grouping

Subtract 4y^{2} from both sides.

$$11y-6-4y^{2}=0$$

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

$$-4y^{2}+11y-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4y^{2}+ay+by-6. To find a and b, set up a system to be solved.

$$a+b=11 ab=-4\left(-6\right)=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 11.

$$a=8 b=3$$

Rewrite -4y^{2}+11y-6 as \left(-4y^{2}+8y\right)+\left(3y-6\right).

$$\left(-4y^{2}+8y\right)+\left(3y-6\right)$$

Factor out 4y in the first and -3 in the second group.

$$4y\left(-y+2\right)-3\left(-y+2\right)$$

Factor out common term -y+2 by using distributive property.

$$\left(-y+2\right)\left(4y-3\right)$$

To find equation solutions, solve -y+2=0 and 4y-3=0.

$$y=2 y=\frac{3}{4}$$

Steps Using the Quadratic Formula

Subtract 4y^{2} from both sides.

$$11y-6-4y^{2}=0$$

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.

$$-4y^{2}+11y-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 11 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 11.

$$y=\frac{-11±\sqrt{121-4\left(-4\right)\left(-6\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$y=\frac{-11±\sqrt{121+16\left(-6\right)}}{2\left(-4\right)}$$

Multiply 16 times -6.

$$y=\frac{-11±\sqrt{121-96}}{2\left(-4\right)}$$

Add 121 to -96.

$$y=\frac{-11±\sqrt{25}}{2\left(-4\right)}$$

Take the square root of 25.

$$y=\frac{-11±5}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{-11±5}{-8}$$

Now solve the equation y=\frac{-11±5}{-8} when ± is plus. Add -11 to 5.

$$y=-\frac{6}{-8}$$

Reduce the fraction \frac{-6}{-8} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{-11±5}{-8} when ± is minus. Subtract 5 from -11.

$$y=-\frac{16}{-8}$$

Divide -16 by -8.

$$y=2$$

The equation is now solved.

$$y=\frac{3}{4} y=2$$

Steps for Completing the Square

Subtract 4y^{2} from both sides.

$$11y-6-4y^{2}=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$11y-4y^{2}=6$$

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.

$$-4y^{2}+11y=6$$

Divide both sides by -4.

$$\frac{-4y^{2}+11y}{-4}=\frac{6}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y^{2}+\frac{11}{-4}y=\frac{6}{-4}$$

Divide 11 by -4.

$$y^{2}-\frac{11}{4}y=\frac{6}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

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

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{11}{4}y+\left(-\frac{11}{8}\right)^{2}=-\frac{3}{2}+\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{11}{4}y+\frac{121}{64}=-\frac{3}{2}+\frac{121}{64}$$

Add -\frac{3}{2} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{11}{4}y+\frac{121}{64}=\frac{25}{64}$$

Factor y^{2}-\frac{11}{4}y+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{11}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{11}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$y-\frac{11}{8}=\frac{5}{8} y-\frac{11}{8}=-\frac{5}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$y=2 y=\frac{3}{4}$$