Solve 7=4+c^2-2c

$$7 = 4 + c ^ { 2 } - 2 c$$

Solve for c

$$c=-1$$

$$c=3$$

Steps Using Factoring

Swap sides so that all variable terms are on the left hand side.

$$4+c^{2}-2c=7$$

Subtract 7 from both sides.

$$4+c^{2}-2c-7=0$$

Subtract 7 from 4 to get -3.

$$-3+c^{2}-2c=0$$

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

$$c^{2}-2c-3=0$$

To solve the equation, factor c^{2}-2c-3 using formula c^{2}+\left(a+b\right)c+ab=\left(c+a\right)\left(c+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite factored expression \left(c+a\right)\left(c+b\right) using the obtained values.

$$\left(c-3\right)\left(c+1\right)$$

To find equation solutions, solve c-3=0 and c+1=0.

$$c=3 c=-1$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$4+c^{2}-2c=7$$

Subtract 7 from both sides.

$$4+c^{2}-2c-7=0$$

Subtract 7 from 4 to get -3.

$$-3+c^{2}-2c=0$$

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

$$c^{2}-2c-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as c^{2}+ac+bc-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite c^{2}-2c-3 as \left(c^{2}-3c\right)+\left(c-3\right).

$$\left(c^{2}-3c\right)+\left(c-3\right)$$

Factor out c in c^{2}-3c.

$$c\left(c-3\right)+c-3$$

Factor out common term c-3 by using distributive property.

$$\left(c-3\right)\left(c+1\right)$$

To find equation solutions, solve c-3=0 and c+1=0.

$$c=3 c=-1$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4+c^{2}-2c=7$$

Subtract 7 from both sides.

$$4+c^{2}-2c-7=0$$

Subtract 7 from 4 to get -3.

$$-3+c^{2}-2c=0$$

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.

$$c^{2}-2c-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

$$c=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$c=\frac{-\left(-2\right)±\sqrt{4+12}}{2}$$

Add 4 to 12.

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

Take the square root of 16.

$$c=\frac{-\left(-2\right)±4}{2}$$

The opposite of -2 is 2.

$$c=\frac{2±4}{2}$$

Now solve the equation c=\frac{2±4}{2} when ± is plus. Add 2 to 4.

$$c=\frac{6}{2}$$

Divide 6 by 2.

$$c=3$$

Now solve the equation c=\frac{2±4}{2} when ± is minus. Subtract 4 from 2.

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

Divide -2 by 2.

$$c=-1$$

The equation is now solved.

$$c=3 c=-1$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4+c^{2}-2c=7$$

Subtract 4 from both sides.

$$c^{2}-2c=7-4$$

Subtract 4 from 7 to get 3.

$$c^{2}-2c=3$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-2c+1=3+1$$

Add 3 to 1.

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

Factor c^{2}-2c+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$c-1=2 c-1=-2$$

Add 1 to both sides of the equation.

$$c=3 c=-1$$