Solve y+2y^2+y=4

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

Solve for y

$$y=-2$$

$$y=1$$

Steps Using Factoring By Grouping

Combine y and y to get 2y.

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

Subtract 4 from both sides.

$$2y+2y^{2}-4=0$$

Divide both sides by 2.

$$y+y^{2}-2=0$$

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

$$y^{2}+y-2=0$$

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

$$a+b=1 ab=1\left(-2\right)=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=2$$

Rewrite y^{2}+y-2 as \left(y^{2}-y\right)+\left(2y-2\right).

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

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

$$y\left(y-1\right)+2\left(y-1\right)$$

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

$$\left(y-1\right)\left(y+2\right)$$

To find equation solutions, solve y-1=0 and y+2=0.

$$y=1 y=-2$$

Steps Using the Quadratic Formula

Combine y and y to get 2y.

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

Subtract 4 from both sides.

$$2y+2y^{2}-4=0$$

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.

$$2y^{2}+2y-4=0$$

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

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

Square 2.

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

Multiply -4 times 2.

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

Multiply -8 times -4.

$$y=\frac{-2±\sqrt{4+32}}{2\times 2}$$

Add 4 to 32.

$$y=\frac{-2±\sqrt{36}}{2\times 2}$$

Take the square root of 36.

$$y=\frac{-2±6}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-2±6}{4}$$

Now solve the equation y=\frac{-2±6}{4} when ± is plus. Add -2 to 6.

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

Divide 4 by 4.

$$y=1$$

Now solve the equation y=\frac{-2±6}{4} when ± is minus. Subtract 6 from -2.

$$y=-\frac{8}{4}$$

Divide -8 by 4.

$$y=-2$$

The equation is now solved.

$$y=1 y=-2$$

Steps for Completing the Square

Combine y and y to get 2y.

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

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.

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

Divide both sides by 2.

$$\frac{2y^{2}+2y}{2}=\frac{4}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\frac{2}{2}y=\frac{4}{2}$$

Divide 2 by 2.

$$y^{2}+y=\frac{4}{2}$$

Divide 4 by 2.

$$y^{2}+y=2$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+y+\frac{1}{4}=2+\frac{1}{4}$$

Add 2 to \frac{1}{4}.

$$y^{2}+y+\frac{1}{4}=\frac{9}{4}$$

Factor y^{2}+y+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{1}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$y+\frac{1}{2}=\frac{3}{2} y+\frac{1}{2}=-\frac{3}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$y=1 y=-2$$