Solve 21y^2=8y+4

$$21 y ^ { 2 } = 8 y + 4$$

Solve for y

$$y=-\frac{2}{7}\approx -0.285714286$$

$$y=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

Subtract 8y from both sides.

$$21y^{2}-8y=4$$

Subtract 4 from both sides.

$$21y^{2}-8y-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 21y^{2}+ay+by-4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=21\left(-4\right)=-84$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -84.

$$1,-84 2,-42 3,-28 4,-21 6,-14 7,-12$$

Calculate the sum for each pair.

$$1-84=-83 2-42=-40 3-28=-25 4-21=-17 6-14=-8 7-12=-5$$

The solution is the pair that gives sum -8.

$$a=-14 b=6$$

Rewrite 21y^{2}-8y-4 as \left(21y^{2}-14y\right)+\left(6y-4\right).

$$\left(21y^{2}-14y\right)+\left(6y-4\right)$$

Factor out 7y in the first and 2 in the second group.

$$7y\left(3y-2\right)+2\left(3y-2\right)$$

Factor out common term 3y-2 by using distributive property.

$$\left(3y-2\right)\left(7y+2\right)$$

To find equation solutions, solve 3y-2=0 and 7y+2=0.

$$y=\frac{2}{3} y=-\frac{2}{7}$$

Steps Using the Quadratic Formula

Subtract 8y from both sides.

$$21y^{2}-8y=4$$

Subtract 4 from both sides.

$$21y^{2}-8y-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 21\left(-4\right)}}{2\times 21}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 21\left(-4\right)}}{2\times 21}$$

Multiply -4 times 21.

$$y=\frac{-\left(-8\right)±\sqrt{64-84\left(-4\right)}}{2\times 21}$$

Multiply -84 times -4.

$$y=\frac{-\left(-8\right)±\sqrt{64+336}}{2\times 21}$$

Add 64 to 336.

$$y=\frac{-\left(-8\right)±\sqrt{400}}{2\times 21}$$

Take the square root of 400.

$$y=\frac{-\left(-8\right)±20}{2\times 21}$$

The opposite of -8 is 8.

$$y=\frac{8±20}{2\times 21}$$

Multiply 2 times 21.

$$y=\frac{8±20}{42}$$

Now solve the equation y=\frac{8±20}{42} when ± is plus. Add 8 to 20.

$$y=\frac{28}{42}$$

Reduce the fraction \frac{28}{42} to lowest terms by extracting and canceling out 14.

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

Now solve the equation y=\frac{8±20}{42} when ± is minus. Subtract 20 from 8.

$$y=-\frac{12}{42}$$

Reduce the fraction \frac{-12}{42} to lowest terms by extracting and canceling out 6.

$$y=-\frac{2}{7}$$

The equation is now solved.

$$y=\frac{2}{3} y=-\frac{2}{7}$$

Steps for Completing the Square

Subtract 8y from both sides.

$$21y^{2}-8y=4$$

Divide both sides by 21.

$$\frac{21y^{2}-8y}{21}=\frac{4}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$y^{2}-\frac{8}{21}y=\frac{4}{21}$$

Divide -\frac{8}{21}, the coefficient of the x term, by 2 to get -\frac{4}{21}. Then add the square of -\frac{4}{21} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{8}{21}y+\left(-\frac{4}{21}\right)^{2}=\frac{4}{21}+\left(-\frac{4}{21}\right)^{2}$$

Square -\frac{4}{21} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{8}{21}y+\frac{16}{441}=\frac{4}{21}+\frac{16}{441}$$

Add \frac{4}{21} to \frac{16}{441} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{8}{21}y+\frac{16}{441}=\frac{100}{441}$$

Factor y^{2}-\frac{8}{21}y+\frac{16}{441}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{4}{21}\right)^{2}=\frac{100}{441}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{4}{21}\right)^{2}}=\sqrt{\frac{100}{441}}$$

Simplify.

$$y-\frac{4}{21}=\frac{10}{21} y-\frac{4}{21}=-\frac{10}{21}$$

Add \frac{4}{21} to both sides of the equation.

$$y=\frac{2}{3} y=-\frac{2}{7}$$