Solve 5y^2+4=-9y

$$5 y ^ { 2 } + 4 = - 9 y$$

Solve for y

$$y=-\frac{4}{5}=-0.8$$

$$y=-1$$

Steps Using Factoring By Grouping

Add 9y to both sides.

$$5y^{2}+4+9y=0$$

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

$$5y^{2}+9y+4=0$$

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.

$$a+b=9 ab=5\times 4=20$$

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 20.

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

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 9.

$$a=4 b=5$$

Rewrite 5y^{2}+9y+4 as \left(5y^{2}+4y\right)+\left(5y+4\right).

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

Factor out y in 5y^{2}+4y.

$$y\left(5y+4\right)+5y+4$$

Factor out common term 5y+4 by using distributive property.

$$\left(5y+4\right)\left(y+1\right)$$

To find equation solutions, solve 5y+4=0 and y+1=0.

$$y=-\frac{4}{5} y=-1$$

Steps Using the Quadratic Formula

Add 9y to both sides.

$$5y^{2}+4+9y=0$$

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.

$$5y^{2}+9y+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 9 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-9±\sqrt{9^{2}-4\times 5\times 4}}{2\times 5}$$

Square 9.

$$y=\frac{-9±\sqrt{81-4\times 5\times 4}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-9±\sqrt{81-20\times 4}}{2\times 5}$$

Multiply -20 times 4.

$$y=\frac{-9±\sqrt{81-80}}{2\times 5}$$

Add 81 to -80.

$$y=\frac{-9±\sqrt{1}}{2\times 5}$$

Take the square root of 1.

$$y=\frac{-9±1}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{-9±1}{10}$$

Now solve the equation y=\frac{-9±1}{10} when ± is plus. Add -9 to 1.

$$y=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{-9±1}{10} when ± is minus. Subtract 1 from -9.

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

Divide -10 by 10.

$$y=-1$$

The equation is now solved.

$$y=-\frac{4}{5} y=-1$$

Steps for Completing the Square

Add 9y to both sides.

$$5y^{2}+4+9y=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$5y^{2}+9y=-4$$

Divide both sides by 5.

$$\frac{5y^{2}+9y}{5}=-\frac{4}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\frac{9}{5}y=-\frac{4}{5}$$

Divide \frac{9}{5}, the coefficient of the x term, by 2 to get \frac{9}{10}. Then add the square of \frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{9}{5}y+\left(\frac{9}{10}\right)^{2}=-\frac{4}{5}+\left(\frac{9}{10}\right)^{2}$$

Square \frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{9}{5}y+\frac{81}{100}=-\frac{4}{5}+\frac{81}{100}$$

Add -\frac{4}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{9}{5}y+\frac{81}{100}=\frac{1}{100}$$

Factor y^{2}+\frac{9}{5}y+\frac{81}{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}.

$$\left(y+\frac{9}{10}\right)^{2}=\frac{1}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{9}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Simplify.

$$y+\frac{9}{10}=\frac{1}{10} y+\frac{9}{10}=-\frac{1}{10}$$

Subtract \frac{9}{10} from both sides of the equation.

$$y=-\frac{4}{5} y=-1$$