Solve 5y^2-8y=4

$$5 y ^ { 2 } - 8 y = 4$$

Solve for y

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

$$y=2$$

Steps Using Factoring By Grouping

Subtract 4 from both sides.

$$5y^{2}-8y-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5y^{2}+ay+by-4. To find a and b, set up a system to be solved.

$$a+b=-8 ab=5\left(-4\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=-10 b=2$$

Rewrite 5y^{2}-8y-4 as \left(5y^{2}-10y\right)+\left(2y-4\right).

$$\left(5y^{2}-10y\right)+\left(2y-4\right)$$

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

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

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

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

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

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

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.

$$5y^{2}-8y=4$$

Subtract 4 from both sides of the equation.

$$5y^{2}-8y-4=4-4$$

Subtracting 4 from itself leaves 0.

$$5y^{2}-8y-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\left(-4\right)}}{2\times 5}$$

Square -8.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 5\left(-4\right)}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-8\right)±\sqrt{64-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$y=\frac{-\left(-8\right)±\sqrt{64+80}}{2\times 5}$$

Add 64 to 80.

$$y=\frac{-\left(-8\right)±\sqrt{144}}{2\times 5}$$

Take the square root of 144.

$$y=\frac{-\left(-8\right)±12}{2\times 5}$$

The opposite of -8 is 8.

$$y=\frac{8±12}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{8±12}{10}$$

Now solve the equation y=\frac{8±12}{10} when ± is plus. Add 8 to 12.

$$y=\frac{20}{10}$$

Divide 20 by 10.

$$y=2$$

Now solve the equation y=\frac{8±12}{10} when ± is minus. Subtract 12 from 8.

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

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

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

The equation is now solved.

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

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.

$$5y^{2}-8y=4$$

Divide both sides by 5.

$$\frac{5y^{2}-8y}{5}=\frac{4}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}-\frac{8}{5}y=\frac{4}{5}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{8}{5}y+\left(-\frac{4}{5}\right)^{2}=\frac{4}{5}+\left(-\frac{4}{5}\right)^{2}$$

Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{8}{5}y+\frac{16}{25}=\frac{4}{5}+\frac{16}{25}$$

Add \frac{4}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{8}{5}y+\frac{16}{25}=\frac{36}{25}$$

Factor y^{2}-\frac{8}{5}y+\frac{16}{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}.

$$\left(y-\frac{4}{5}\right)^{2}=\frac{36}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{4}{5}\right)^{2}}=\sqrt{\frac{36}{25}}$$

Simplify.

$$y-\frac{4}{5}=\frac{6}{5} y-\frac{4}{5}=-\frac{6}{5}$$

Add \frac{4}{5} to both sides of the equation.

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