Solve y+3+y^2-1=4y+6

$$y+3+ { y }^{ 2 } -1=4y+6$$

Solve for y

$$y=-1$$

$$y=4$$

Steps Using Factoring

Subtract 1 from 3 to get 2.

$$y+2+y^{2}=4y+6$$

Subtract 4y from both sides.

$$y+2+y^{2}-4y=6$$

Combine y and -4y to get -3y.

$$-3y+2+y^{2}=6$$

Subtract 6 from both sides.

$$-3y+2+y^{2}-6=0$$

Subtract 6 from 2 to get -4.

$$-3y-4+y^{2}=0$$

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

$$y^{2}-3y-4=0$$

To solve the equation, factor y^{2}-3y-4 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-3 ab=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

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

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

$$y=4 y=-1$$

Steps Using Factoring By Grouping

Subtract 1 from 3 to get 2.

$$y+2+y^{2}=4y+6$$

Subtract 4y from both sides.

$$y+2+y^{2}-4y=6$$

Combine y and -4y to get -3y.

$$-3y+2+y^{2}=6$$

Subtract 6 from both sides.

$$-3y+2+y^{2}-6=0$$

Subtract 6 from 2 to get -4.

$$-3y-4+y^{2}=0$$

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

$$y^{2}-3y-4=0$$

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

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite y^{2}-3y-4 as \left(y^{2}-4y\right)+\left(y-4\right).

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

Factor out y in y^{2}-4y.

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

Factor out common term y-4 by using distributive property.

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

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

$$y=4 y=-1$$

Steps Using the Quadratic Formula

Subtract 1 from 3 to get 2.

$$y+2+y^{2}=4y+6$$

Subtract 4y from both sides.

$$y+2+y^{2}-4y=6$$

Combine y and -4y to get -3y.

$$-3y+2+y^{2}=6$$

Subtract 6 from both sides.

$$-3y+2+y^{2}-6=0$$

Subtract 6 from 2 to get -4.

$$-3y-4+y^{2}=0$$

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.

$$y^{2}-3y-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)}}{2}$$

Square -3.

$$y=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$y=\frac{-\left(-3\right)±\sqrt{9+16}}{2}$$

Add 9 to 16.

$$y=\frac{-\left(-3\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$y=\frac{-\left(-3\right)±5}{2}$$

The opposite of -3 is 3.

$$y=\frac{3±5}{2}$$

Now solve the equation y=\frac{3±5}{2} when ± is plus. Add 3 to 5.

$$y=\frac{8}{2}$$

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{3±5}{2} when ± is minus. Subtract 5 from 3.

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

Divide -2 by 2.

$$y=-1$$

The equation is now solved.

$$y=4 y=-1$$

Steps for Completing the Square

Subtract 1 from 3 to get 2.

$$y+2+y^{2}=4y+6$$

Subtract 4y from both sides.

$$y+2+y^{2}-4y=6$$

Combine y and -4y to get -3y.

$$-3y+2+y^{2}=6$$

Subtract 2 from both sides.

$$-3y+y^{2}=6-2$$

Subtract 2 from 6 to get 4.

$$-3y+y^{2}=4$$

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.

$$y^{2}-3y=4$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-3y+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-3y+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$y^{2}-3y+\frac{9}{4}=\frac{25}{4}$$

Factor y^{2}-3y+\frac{9}{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}.

$$\left(y-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$y=4 y=-1$$