Solve 16(x-4)-(x+3)=(x+3)(x-4)

$$16 ( x - 4 ) - ( x + 3 ) = ( x + 3 ) ( x - 4 )$$

Solve for x

$$x=5$$

$$x=11$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 16 by x-4.

$$16x-64-\left(x+3\right)=\left(x+3\right)\left(x-4\right)$$

To find the opposite of x+3, find the opposite of each term.

$$16x-64-x-3=\left(x+3\right)\left(x-4\right)$$

Combine 16x and -x to get 15x.

$$15x-64-3=\left(x+3\right)\left(x-4\right)$$

Subtract 3 from -64 to get -67.

$$15x-67=\left(x+3\right)\left(x-4\right)$$

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

$$15x-67=x^{2}-x-12$$

Subtract x^{2} from both sides.

$$15x-67-x^{2}=-x-12$$

Add x to both sides.

$$15x-67-x^{2}+x=-12$$

Combine 15x and x to get 16x.

$$16x-67-x^{2}=-12$$

Add 12 to both sides.

$$16x-67-x^{2}+12=0$$

Add -67 and 12 to get -55.

$$16x-55-x^{2}=0$$

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.

$$-x^{2}+16x-55=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 16 for b, and -55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-1\right)\left(-55\right)}}{2\left(-1\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-1\right)\left(-55\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-16±\sqrt{256+4\left(-55\right)}}{2\left(-1\right)}$$

Multiply 4 times -55.

$$x=\frac{-16±\sqrt{256-220}}{2\left(-1\right)}$$

Add 256 to -220.

$$x=\frac{-16±\sqrt{36}}{2\left(-1\right)}$$

Take the square root of 36.

$$x=\frac{-16±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-16±6}{-2}$$

Now solve the equation x=\frac{-16±6}{-2} when ± is plus. Add -16 to 6.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

Now solve the equation x=\frac{-16±6}{-2} when ± is minus. Subtract 6 from -16.

$$x=-\frac{22}{-2}$$

Divide -22 by -2.

$$x=11$$

The equation is now solved.

$$x=5 x=11$$

Steps for Completing the Square

Use the distributive property to multiply 16 by x-4.

$$16x-64-\left(x+3\right)=\left(x+3\right)\left(x-4\right)$$

To find the opposite of x+3, find the opposite of each term.

$$16x-64-x-3=\left(x+3\right)\left(x-4\right)$$

Combine 16x and -x to get 15x.

$$15x-64-3=\left(x+3\right)\left(x-4\right)$$

Subtract 3 from -64 to get -67.

$$15x-67=\left(x+3\right)\left(x-4\right)$$

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

$$15x-67=x^{2}-x-12$$

Subtract x^{2} from both sides.

$$15x-67-x^{2}=-x-12$$

Add x to both sides.

$$15x-67-x^{2}+x=-12$$

Combine 15x and x to get 16x.

$$16x-67-x^{2}=-12$$

Add 67 to both sides.

$$16x-x^{2}=-12+67$$

Add -12 and 67 to get 55.

$$16x-x^{2}=55$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+16x=55$$

Divide both sides by -1.

$$\frac{-x^{2}+16x}{-1}=\frac{55}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{16}{-1}x=\frac{55}{-1}$$

Divide 16 by -1.

$$x^{2}-16x=\frac{55}{-1}$$

Divide 55 by -1.

$$x^{2}-16x=-55$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-55+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-55+64$$

Add -55 to 64.

$$x^{2}-16x+64=9$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-8\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-8=3 x-8=-3$$

Add 8 to both sides of the equation.

$$x=11 x=5$$