Solve (x-3)(x+4)+8=x

$$( x - 3 ) ( x + 4 ) + 8 = x$$

Solve for x

$$x=-2$$

$$x=2$$

Steps Using Difference of Squares

Use the distributive property to multiply x-3 by x+4 and combine like terms.

$$x^{2}+x-12+8=x$$

Add -12 and 8 to get -4.

$$x^{2}+x-4=x$$

Subtract x from both sides.

$$x^{2}+x-4-x=0$$

Combine x and -x to get 0.

$$x^{2}-4=0$$

Consider x^{2}-4. Rewrite x^{2}-4 as x^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-2\right)\left(x+2\right)=0$$

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

$$x=2 x=-2$$

Steps by Finding Square Root

Use the distributive property to multiply x-3 by x+4 and combine like terms.

$$x^{2}+x-12+8=x$$

Add -12 and 8 to get -4.

$$x^{2}+x-4=x$$

Subtract x from both sides.

$$x^{2}+x-4-x=0$$

Combine x and -x to get 0.

$$x^{2}-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-3 by x+4 and combine like terms.

$$x^{2}+x-12+8=x$$

Add -12 and 8 to get -4.

$$x^{2}+x-4=x$$

Subtract x from both sides.

$$x^{2}+x-4-x=0$$

Combine x and -x to get 0.

$$x^{2}-4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-4\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{0±4}{2}$$

Now solve the equation x=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=2 x=-2$$