Solve (x-3)^2-4=(5+x)(5-x)

$$( x - 3 ) ^ { 2 } - 4 = ( 5 + x ) ( 5 - x )$$

Solve for x

$$x=-2$$

$$x=5$$

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}.

$$x^{2}-6x+9-4=\left(5+x\right)\left(5-x\right)$$

Subtract 4 from 9 to get 5.

$$x^{2}-6x+5=\left(5+x\right)\left(5-x\right)$$

Consider \left(5+x\right)\left(5-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-6x+5=25-x^{2}$$

Subtract 25 from both sides.

$$x^{2}-6x+5-25=-x^{2}$$

Subtract 25 from 5 to get -20.

$$x^{2}-6x-20=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}-6x-20+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-6x-20=0$$

Divide both sides by 2.

$$x^{2}-3x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-10\right)=-10$$

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 -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

Rewrite x^{2}-3x-10 as \left(x^{2}-5x\right)+\left(2x-10\right).

$$\left(x^{2}-5x\right)+\left(2x-10\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(x-5\right)+2\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

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}.

$$x^{2}-6x+9-4=\left(5+x\right)\left(5-x\right)$$

Subtract 4 from 9 to get 5.

$$x^{2}-6x+5=\left(5+x\right)\left(5-x\right)$$

Consider \left(5+x\right)\left(5-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-6x+5=25-x^{2}$$

Subtract 25 from both sides.

$$x^{2}-6x+5-25=-x^{2}$$

Subtract 25 from 5 to get -20.

$$x^{2}-6x-20=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}-6x-20+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-6x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -6 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-20\right)}}{2\times 2}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-20\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-6\right)±\sqrt{36-8\left(-20\right)}}{2\times 2}$$

Multiply -8 times -20.

$$x=\frac{-\left(-6\right)±\sqrt{36+160}}{2\times 2}$$

Add 36 to 160.

$$x=\frac{-\left(-6\right)±\sqrt{196}}{2\times 2}$$

Take the square root of 196.

$$x=\frac{-\left(-6\right)±14}{2\times 2}$$

The opposite of -6 is 6.

$$x=\frac{6±14}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{6±14}{4}$$

Now solve the equation x=\frac{6±14}{4} when ± is plus. Add 6 to 14.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{6±14}{4} when ± is minus. Subtract 14 from 6.

$$x=-\frac{8}{4}$$

Divide -8 by 4.

$$x=-2$$

The equation is now solved.

$$x=5 x=-2$$

Steps for Completing the Square

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

$$x^{2}-6x+9-4=\left(5+x\right)\left(5-x\right)$$

Subtract 4 from 9 to get 5.

$$x^{2}-6x+5=\left(5+x\right)\left(5-x\right)$$

Consider \left(5+x\right)\left(5-x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-6x+5=25-x^{2}$$

Add x^{2} to both sides.

$$x^{2}-6x+5+x^{2}=25$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-6x+5=25$$

Subtract 5 from both sides.

$$2x^{2}-6x=25-5$$

Subtract 5 from 25 to get 20.

$$2x^{2}-6x=20$$

Divide both sides by 2.

$$\frac{2x^{2}-6x}{2}=\frac{20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{6}{2}\right)x=\frac{20}{2}$$

Divide -6 by 2.

$$x^{2}-3x=\frac{20}{2}$$

Divide 20 by 2.

$$x^{2}-3x=10$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=10+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=10+\frac{9}{4}$$

Add 10 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{49}{4}$$

Factor x^{2}-3x+\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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{7}{2} x-\frac{3}{2}=-\frac{7}{2}$$

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

$$x=5 x=-2$$