Solve 8-26y+11y^2=0

$$8 - 26 y + 11 y ^ { 2 } = 0$$

Solve for y

$$y=2$$

$$y=\frac{4}{11}\approx 0.363636364$$

Steps Using Factoring By Grouping

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

$$11y^{2}-26y+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 11y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=-26 ab=11\times 8=88$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 88.

$$-1,-88 -2,-44 -4,-22 -8,-11$$

Calculate the sum for each pair.

$$-1-88=-89 -2-44=-46 -4-22=-26 -8-11=-19$$

The solution is the pair that gives sum -26.

$$a=-22 b=-4$$

Rewrite 11y^{2}-26y+8 as \left(11y^{2}-22y\right)+\left(-4y+8\right).

$$\left(11y^{2}-22y\right)+\left(-4y+8\right)$$

Factor out 11y in the first and -4 in the second group.

$$11y\left(y-2\right)-4\left(y-2\right)$$

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

$$\left(y-2\right)\left(11y-4\right)$$

To find equation solutions, solve y-2=0 and 11y-4=0.

$$y=2 y=\frac{4}{11}$$

Steps Using the Quadratic Formula

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.

$$11y^{2}-26y+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 11 for a, -26 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\times 11\times 8}}{2\times 11}$$

Square -26.

$$y=\frac{-\left(-26\right)±\sqrt{676-4\times 11\times 8}}{2\times 11}$$

Multiply -4 times 11.

$$y=\frac{-\left(-26\right)±\sqrt{676-44\times 8}}{2\times 11}$$

Multiply -44 times 8.

$$y=\frac{-\left(-26\right)±\sqrt{676-352}}{2\times 11}$$

Add 676 to -352.

$$y=\frac{-\left(-26\right)±\sqrt{324}}{2\times 11}$$

Take the square root of 324.

$$y=\frac{-\left(-26\right)±18}{2\times 11}$$

The opposite of -26 is 26.

$$y=\frac{26±18}{2\times 11}$$

Multiply 2 times 11.

$$y=\frac{26±18}{22}$$

Now solve the equation y=\frac{26±18}{22} when ± is plus. Add 26 to 18.

$$y=\frac{44}{22}$$

Divide 44 by 22.

$$y=2$$

Now solve the equation y=\frac{26±18}{22} when ± is minus. Subtract 18 from 26.

$$y=\frac{8}{22}$$

Reduce the fraction \frac{8}{22} to lowest terms by extracting and canceling out 2.

$$y=\frac{4}{11}$$

The equation is now solved.

$$y=2 y=\frac{4}{11}$$

Steps for Completing the Square

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.

$$11y^{2}-26y+8=0$$

Subtract 8 from both sides of the equation.

$$11y^{2}-26y+8-8=-8$$

Subtracting 8 from itself leaves 0.

$$11y^{2}-26y=-8$$

Divide both sides by 11.

$$\frac{11y^{2}-26y}{11}=-\frac{8}{11}$$

Dividing by 11 undoes the multiplication by 11.

$$y^{2}-\frac{26}{11}y=-\frac{8}{11}$$

Divide -\frac{26}{11}, the coefficient of the x term, by 2 to get -\frac{13}{11}. Then add the square of -\frac{13}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{26}{11}y+\left(-\frac{13}{11}\right)^{2}=-\frac{8}{11}+\left(-\frac{13}{11}\right)^{2}$$

Square -\frac{13}{11} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{26}{11}y+\frac{169}{121}=-\frac{8}{11}+\frac{169}{121}$$

Add -\frac{8}{11} to \frac{169}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{26}{11}y+\frac{169}{121}=\frac{81}{121}$$

Factor y^{2}-\frac{26}{11}y+\frac{169}{121}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{13}{11}\right)^{2}=\frac{81}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{13}{11}\right)^{2}}=\sqrt{\frac{81}{121}}$$

Simplify.

$$y-\frac{13}{11}=\frac{9}{11} y-\frac{13}{11}=-\frac{9}{11}$$

Add \frac{13}{11} to both sides of the equation.

$$y=2 y=\frac{4}{11}$$