Solve y^2+6y+9=49

$$y ^ { 2 } + 6 y + 9 = 49$$

Solve for y

$$y=-10$$

$$y=4$$

Steps Using Factoring

Subtract 49 from both sides.

$$y^{2}+6y+9-49=0$$

Subtract 49 from 9 to get -40.

$$y^{2}+6y-40=0$$

To solve the equation, factor y^{2}+6y-40 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.

$$a+b=6 ab=-40$$

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. List all such integer pairs that give product -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 6.

$$a=-4 b=10$$

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

$$\left(y-4\right)\left(y+10\right)$$

To find equation solutions, solve y-4=0 and y+10=0.

$$y=4 y=-10$$

Steps Using Factoring By Grouping

Subtract 49 from both sides.

$$y^{2}+6y+9-49=0$$

Subtract 49 from 9 to get -40.

$$y^{2}+6y-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-40. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-40\right)=-40$$

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. List all such integer pairs that give product -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 6.

$$a=-4 b=10$$

Rewrite y^{2}+6y-40 as \left(y^{2}-4y\right)+\left(10y-40\right).

$$\left(y^{2}-4y\right)+\left(10y-40\right)$$

Factor out y in the first and 10 in the second group.

$$y\left(y-4\right)+10\left(y-4\right)$$

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

$$\left(y-4\right)\left(y+10\right)$$

To find equation solutions, solve y-4=0 and y+10=0.

$$y=4 y=-10$$

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.

$$y^{2}+6y+9=49$$

Subtract 49 from both sides of the equation.

$$y^{2}+6y+9-49=49-49$$

Subtracting 49 from itself leaves 0.

$$y^{2}+6y+9-49=0$$

Subtract 49 from 9.

$$y^{2}+6y-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-6±\sqrt{6^{2}-4\left(-40\right)}}{2}$$

Square 6.

$$y=\frac{-6±\sqrt{36-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$y=\frac{-6±\sqrt{36+160}}{2}$$

Add 36 to 160.

$$y=\frac{-6±\sqrt{196}}{2}$$

Take the square root of 196.

$$y=\frac{-6±14}{2}$$

Now solve the equation y=\frac{-6±14}{2} when ± is plus. Add -6 to 14.

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

Divide 8 by 2.

$$y=4$$

Now solve the equation y=\frac{-6±14}{2} when ± is minus. Subtract 14 from -6.

$$y=-\frac{20}{2}$$

Divide -20 by 2.

$$y=-10$$

The equation is now solved.

$$y=4 y=-10$$

Steps for Completing the Square

Factor y^{2}+6y+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+3\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+3\right)^{2}}=\sqrt{49}$$

Simplify.

$$y+3=7 y+3=-7$$

Subtract 3 from both sides of the equation.

$$y=4 y=-10$$