Solve 4c=16+c^2-12

$$4 c = 16 + c ^ { 2 } - 12$$

Solve for c

$$c=2$$

Steps Using Factoring By Grouping

Subtract 12 from 16 to get 4.

$$4c=4+c^{2}$$

Subtract 4 from both sides.

$$4c-4=c^{2}$$

Subtract c^{2} from both sides.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -c^{2}+ac+bc-4. To find a and b, set up a system to be solved.

$$a+b=4 ab=-\left(-4\right)=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 4.

$$a=2 b=2$$

Rewrite -c^{2}+4c-4 as \left(-c^{2}+2c\right)+\left(2c-4\right).

$$\left(-c^{2}+2c\right)+\left(2c-4\right)$$

Factor out -c in the first and 2 in the second group.

$$-c\left(c-2\right)+2\left(c-2\right)$$

Factor out common term c-2 by using distributive property.

$$\left(c-2\right)\left(-c+2\right)$$

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

$$c=2 c=2$$

Steps Using the Quadratic Formula

Subtract 12 from 16 to get 4.

$$4c=4+c^{2}$$

Subtract 4 from both sides.

$$4c-4=c^{2}$$

Subtract c^{2} from both sides.

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

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.

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

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

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

Square 4.

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

Multiply -4 times -1.

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

Multiply 4 times -4.

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

Add 16 to -16.

$$c=\frac{-4±\sqrt{0}}{2\left(-1\right)}$$

Take the square root of 0.

$$c=-\frac{4}{2\left(-1\right)}$$

Multiply 2 times -1.

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

Divide -4 by -2.

$$c=2$$

Steps for Completing the Square

Subtract 12 from 16 to get 4.

$$4c=4+c^{2}$$

Subtract c^{2} from both sides.

$$4c-c^{2}=4$$

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.

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

Divide both sides by -1.

$$\frac{-c^{2}+4c}{-1}=\frac{4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$c^{2}+\frac{4}{-1}c=\frac{4}{-1}$$

Divide 4 by -1.

$$c^{2}-4c=\frac{4}{-1}$$

Divide 4 by -1.

$$c^{2}-4c=-4$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-4c+\left(-2\right)^{2}=-4+\left(-2\right)^{2}$$

Square -2.

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

Add -4 to 4.

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

Factor c^{2}-4c+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}.

$$\left(c-2\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-2\right)^{2}}=\sqrt{0}$$

Simplify.

$$c-2=0 c-2=0$$

Add 2 to both sides of the equation.

$$c=2 c=2$$

The equation is now solved. Solutions are the same.

$$c=2$$