Solve 5y^2+22y+17=0

$$5 { y }^{ 2 } +22y+17=0$$

Solve for y

$$y = -\frac{17}{5} = -3\frac{2}{5} = -3.4$$

$$y=-1$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5y^{2}+ay+by+17. To find a and b, set up a system to be solved.

$$a+b=22 ab=5\times 17=85$$

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 85.

$$1,85 5,17$$

Calculate the sum for each pair.

$$1+85=86 5+17=22$$

The solution is the pair that gives sum 22.

$$a=5 b=17$$

Rewrite 5y^{2}+22y+17 as \left(5y^{2}+5y\right)+\left(17y+17\right).

$$\left(5y^{2}+5y\right)+\left(17y+17\right)$$

Factor out 5y in the first and 17 in the second group.

$$5y\left(y+1\right)+17\left(y+1\right)$$

Factor out common term y+1 by using distributive property.

$$\left(y+1\right)\left(5y+17\right)$$

To find equation solutions, solve y+1=0 and 5y+17=0.

$$y=-1 y=-\frac{17}{5}$$

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.

$$5y^{2}+22y+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 22 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-22±\sqrt{22^{2}-4\times 5\times 17}}{2\times 5}$$

Square 22.

$$y=\frac{-22±\sqrt{484-4\times 5\times 17}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-22±\sqrt{484-20\times 17}}{2\times 5}$$

Multiply -20 times 17.

$$y=\frac{-22±\sqrt{484-340}}{2\times 5}$$

Add 484 to -340.

$$y=\frac{-22±\sqrt{144}}{2\times 5}$$

Take the square root of 144.

$$y=\frac{-22±12}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{-22±12}{10}$$

Now solve the equation y=\frac{-22±12}{10} when ± is plus. Add -22 to 12.

$$y=-\frac{10}{10}$$

Divide -10 by 10.

$$y=-1$$

Now solve the equation y=\frac{-22±12}{10} when ± is minus. Subtract 12 from -22.

$$y=-\frac{34}{10}$$

Reduce the fraction \frac{-34}{10} to lowest terms by extracting and canceling out 2.

$$y=-\frac{17}{5}$$

The equation is now solved.

$$y=-1 y=-\frac{17}{5}$$

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.

$$5y^{2}+22y+17=0$$

Subtract 17 from both sides of the equation.

$$5y^{2}+22y+17-17=-17$$

Subtracting 17 from itself leaves 0.

$$5y^{2}+22y=-17$$

Divide both sides by 5.

$$\frac{5y^{2}+22y}{5}=-\frac{17}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\frac{22}{5}y=-\frac{17}{5}$$

Divide \frac{22}{5}, the coefficient of the x term, by 2 to get \frac{11}{5}. Then add the square of \frac{11}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{22}{5}y+\left(\frac{11}{5}\right)^{2}=-\frac{17}{5}+\left(\frac{11}{5}\right)^{2}$$

Square \frac{11}{5} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{22}{5}y+\frac{121}{25}=-\frac{17}{5}+\frac{121}{25}$$

Add -\frac{17}{5} to \frac{121}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{22}{5}y+\frac{121}{25}=\frac{36}{25}$$

Factor y^{2}+\frac{22}{5}y+\frac{121}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{11}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{11}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$y+\frac{11}{5}=\frac{6}{5} y+\frac{11}{5}=-\frac{6}{5}$$

Subtract \frac{11}{5} from both sides of the equation.

$$y=-1 y=-\frac{17}{5}$$