Solve 5y^2=17y-6

$$5 y ^ { 2 } = 17 y - 6$$

Solve for y

$$y=\frac{2}{5}=0.4$$

$$y=3$$

Steps Using Factoring By Grouping

Subtract 17y from both sides.

$$5y^{2}-17y=-6$$

Add 6 to both sides.

$$5y^{2}-17y+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-17 ab=5\times 6=30$$

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 30.

$$-1,-30 -2,-15 -3,-10 -5,-6$$

Calculate the sum for each pair.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

The solution is the pair that gives sum -17.

$$a=-15 b=-2$$

Rewrite 5y^{2}-17y+6 as \left(5y^{2}-15y\right)+\left(-2y+6\right).

$$\left(5y^{2}-15y\right)+\left(-2y+6\right)$$

Factor out 5y in the first and -2 in the second group.

$$5y\left(y-3\right)-2\left(y-3\right)$$

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(5y-2\right)$$

To find equation solutions, solve y-3=0 and 5y-2=0.

$$y=3 y=\frac{2}{5}$$

Steps Using the Quadratic Formula

Subtract 17y from both sides.

$$5y^{2}-17y=-6$$

Add 6 to both sides.

$$5y^{2}-17y+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -17 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 5\times 6}}{2\times 5}$$

Square -17.

$$y=\frac{-\left(-17\right)±\sqrt{289-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-17\right)±\sqrt{289-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$y=\frac{-\left(-17\right)±\sqrt{289-120}}{2\times 5}$$

Add 289 to -120.

$$y=\frac{-\left(-17\right)±\sqrt{169}}{2\times 5}$$

Take the square root of 169.

$$y=\frac{-\left(-17\right)±13}{2\times 5}$$

The opposite of -17 is 17.

$$y=\frac{17±13}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{17±13}{10}$$

Now solve the equation y=\frac{17±13}{10} when ± is plus. Add 17 to 13.

$$y=\frac{30}{10}$$

Divide 30 by 10.

$$y=3$$

Now solve the equation y=\frac{17±13}{10} when ± is minus. Subtract 13 from 17.

$$y=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$y=\frac{2}{5}$$

The equation is now solved.

$$y=3 y=\frac{2}{5}$$

Steps for Completing the Square

Subtract 17y from both sides.

$$5y^{2}-17y=-6$$

Divide both sides by 5.

$$\frac{5y^{2}-17y}{5}=-\frac{6}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}-\frac{17}{5}y=-\frac{6}{5}$$

Divide -\frac{17}{5}, the coefficient of the x term, by 2 to get -\frac{17}{10}. Then add the square of -\frac{17}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{17}{5}y+\left(-\frac{17}{10}\right)^{2}=-\frac{6}{5}+\left(-\frac{17}{10}\right)^{2}$$

Square -\frac{17}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{17}{5}y+\frac{289}{100}=-\frac{6}{5}+\frac{289}{100}$$

Add -\frac{6}{5} to \frac{289}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{17}{5}y+\frac{289}{100}=\frac{169}{100}$$

Factor y^{2}-\frac{17}{5}y+\frac{289}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{17}{10}\right)^{2}=\frac{169}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{17}{10}\right)^{2}}=\sqrt{\frac{169}{100}}$$

Simplify.

$$y-\frac{17}{10}=\frac{13}{10} y-\frac{17}{10}=-\frac{13}{10}$$

Add \frac{17}{10} to both sides of the equation.

$$y=3 y=\frac{2}{5}$$