Solve (x+5)^2+3=19

$$( x + 5 ) ^ { 2 } + 3 = 19$$

Solve for x

$$x=-1$$

$$x=-9$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25+3=19$$

Add 25 and 3 to get 28.

$$x^{2}+10x+28=19$$

Subtract 19 from both sides.

$$x^{2}+10x+28-19=0$$

Subtract 19 from 28 to get 9.

$$x^{2}+10x+9=0$$

To solve the equation, factor x^{2}+10x+9 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.

$$a+b=10 ab=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

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

$$\left(x+1\right)\left(x+9\right)$$

To find equation solutions, solve x+1=0 and x+9=0.

$$x=-1 x=-9$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25+3=19$$

Add 25 and 3 to get 28.

$$x^{2}+10x+28=19$$

Subtract 19 from both sides.

$$x^{2}+10x+28-19=0$$

Subtract 19 from 28 to get 9.

$$x^{2}+10x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 9=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

Rewrite x^{2}+10x+9 as \left(x^{2}+x\right)+\left(9x+9\right).

$$\left(x^{2}+x\right)+\left(9x+9\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x+1\right)+9\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+9\right)$$

To find equation solutions, solve x+1=0 and x+9=0.

$$x=-1 x=-9$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25+3=19$$

Add 25 and 3 to get 28.

$$x^{2}+10x+28=19$$

Subtract 19 from both sides.

$$x^{2}+10x+28-19=0$$

Subtract 19 from 28 to get 9.

$$x^{2}+10x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 10 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 9}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-10±\sqrt{100-36}}{2}$$

Add 100 to -36.

$$x=\frac{-10±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-10±8}{2}$$

Now solve the equation x=\frac{-10±8}{2} when ± is plus. Add -10 to 8.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-10±8}{2} when ± is minus. Subtract 8 from -10.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=-1 x=-9$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25+3=19$$

Add 25 and 3 to get 28.

$$x^{2}+10x+28=19$$

Subtract 28 from both sides.

$$x^{2}+10x=19-28$$

Subtract 28 from 19 to get -9.

$$x^{2}+10x=-9$$

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

$$x^{2}+10x+5^{2}=-9+5^{2}$$

Square 5.

$$x^{2}+10x+25=-9+25$$

Add -9 to 25.

$$x^{2}+10x+25=16$$

Factor x^{2}+10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x+5=4 x+5=-4$$

Subtract 5 from both sides of the equation.

$$x=-1 x=-9$$