Solve 49-(x+5)^2=0

$$49 - ( x + 5 ) ^ { 2 } = 0$$

Solve for x

$$x=2$$

$$x=-12$$

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}.

$$49-\left(x^{2}+10x+25\right)=0$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$49-x^{2}-10x-25=0$$

Subtract 25 from 49 to get 24.

$$24-x^{2}-10x=0$$

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

$$-x^{2}-10x+24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-24=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=2 b=-12$$

Rewrite -x^{2}-10x+24 as \left(-x^{2}+2x\right)+\left(-12x+24\right).

$$\left(-x^{2}+2x\right)+\left(-12x+24\right)$$

Factor out x in the first and 12 in the second group.

$$x\left(-x+2\right)+12\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(x+12\right)$$

To find equation solutions, solve -x+2=0 and x+12=0.

$$x=2 x=-12$$

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}.

$$49-\left(x^{2}+10x+25\right)=0$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$49-x^{2}-10x-25=0$$

Subtract 25 from 49 to get 24.

$$24-x^{2}-10x=0$$

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.

$$-x^{2}-10x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -10 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-1\right)\times 24}}{2\left(-1\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-1\right)\times 24}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-10\right)±\sqrt{100+4\times 24}}{2\left(-1\right)}$$

Multiply 4 times 24.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\left(-1\right)}$$

Add 100 to 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\left(-1\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-10\right)±14}{2\left(-1\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{10±14}{-2}$$

Now solve the equation x=\frac{10±14}{-2} when ± is plus. Add 10 to 14.

$$x=\frac{24}{-2}$$

Divide 24 by -2.

$$x=-12$$

Now solve the equation x=\frac{10±14}{-2} when ± is minus. Subtract 14 from 10.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

The equation is now solved.

$$x=-12 x=2$$

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}.

$$49-\left(x^{2}+10x+25\right)=0$$

To find the opposite of x^{2}+10x+25, find the opposite of each term.

$$49-x^{2}-10x-25=0$$

Subtract 25 from 49 to get 24.

$$24-x^{2}-10x=0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}-10x=-24$$

Divide both sides by -1.

$$\frac{-x^{2}-10x}{-1}=-\frac{24}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{10}{-1}\right)x=-\frac{24}{-1}$$

Divide -10 by -1.

$$x^{2}+10x=-\frac{24}{-1}$$

Divide -24 by -1.

$$x^{2}+10x=24$$

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.

$$x^{2}+10x+5^{2}=24+5^{2}$$

Square 5.

$$x^{2}+10x+25=24+25$$

Add 24 to 25.

$$x^{2}+10x+25=49$$

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}.

$$\left(x+5\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{49}$$

Simplify.

$$x+5=7 x+5=-7$$

Subtract 5 from both sides of the equation.

$$x=2 x=-12$$