Solve left(x+3right)^2=49

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

Solve for x

$$x=4$$

$$x=-10$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract 49 from both sides.

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

Subtract 49 from 9 to get -40.

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

To solve the equation, factor x^{2}+6x-40 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+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(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$x=4 x=-10$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract 49 from both sides.

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

Subtract 49 from 9 to get -40.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-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 x^{2}+6x-40 as \left(x^{2}-4x\right)+\left(10x-40\right).

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

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

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

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

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

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

$$x=4 x=-10$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract 49 from both sides.

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

Subtract 49 from 9 to get -40.

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

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}.

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

Square 6.

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

Multiply -4 times -40.

$$x=\frac{-6±\sqrt{36+160}}{2}$$

Add 36 to 160.

$$x=\frac{-6±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-6±14}{2}$$

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

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

Divide 8 by 2.

$$x=4$$

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

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

Divide -20 by 2.

$$x=-10$$

The equation is now solved.

$$x=4 x=-10$$

Steps for Completing the Square

Take the square root of both sides of the equation.

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

Simplify.

$$x+3=7 x+3=-7$$

Subtract 3 from both sides of the equation.

$$x=4 x=-10$$