Solve 3c^2-7c-3c+7=0

$$3 c ^ { 2 } - 7 c - 3 c + 7 = 0$$

Solve for c

$$c=1$$

$$c = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

Steps Using Factoring By Grouping

Combine -7c and -3c to get -10c.

$$3c^{2}-10c+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3c^{2}+ac+bc+7. To find a and b, set up a system to be solved.

$$a+b=-10 ab=3\times 7=21$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-7 b=-3$$

Rewrite 3c^{2}-10c+7 as \left(3c^{2}-7c\right)+\left(-3c+7\right).

$$\left(3c^{2}-7c\right)+\left(-3c+7\right)$$

Factor out c in the first and -1 in the second group.

$$c\left(3c-7\right)-\left(3c-7\right)$$

Factor out common term 3c-7 by using distributive property.

$$\left(3c-7\right)\left(c-1\right)$$

To find equation solutions, solve 3c-7=0 and c-1=0.

$$c=\frac{7}{3} c=1$$

Steps Using the Quadratic Formula

Combine -7c and -3c to get -10c.

$$3c^{2}-10c+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 7}}{2\times 3}$$

Square -10.

$$c=\frac{-\left(-10\right)±\sqrt{100-4\times 3\times 7}}{2\times 3}$$

Multiply -4 times 3.

$$c=\frac{-\left(-10\right)±\sqrt{100-12\times 7}}{2\times 3}$$

Multiply -12 times 7.

$$c=\frac{-\left(-10\right)±\sqrt{100-84}}{2\times 3}$$

Add 100 to -84.

$$c=\frac{-\left(-10\right)±\sqrt{16}}{2\times 3}$$

Take the square root of 16.

$$c=\frac{-\left(-10\right)±4}{2\times 3}$$

The opposite of -10 is 10.

$$c=\frac{10±4}{2\times 3}$$

Multiply 2 times 3.

$$c=\frac{10±4}{6}$$

Now solve the equation c=\frac{10±4}{6} when ± is plus. Add 10 to 4.

$$c=\frac{14}{6}$$

Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

$$c=\frac{7}{3}$$

Now solve the equation c=\frac{10±4}{6} when ± is minus. Subtract 4 from 10.

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

Divide 6 by 6.

$$c=1$$

The equation is now solved.

$$c=\frac{7}{3} c=1$$

Steps for Completing the Square

Combine -7c and -3c to get -10c.

$$3c^{2}-10c+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$3c^{2}-10c=-7$$

Divide both sides by 3.

$$\frac{3c^{2}-10c}{3}=-\frac{7}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$c^{2}-\frac{10}{3}c=-\frac{7}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-\frac{10}{3}c+\left(-\frac{5}{3}\right)^{2}=-\frac{7}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$c^{2}-\frac{10}{3}c+\frac{25}{9}=-\frac{7}{3}+\frac{25}{9}$$

Add -\frac{7}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$c^{2}-\frac{10}{3}c+\frac{25}{9}=\frac{4}{9}$$

Factor c^{2}-\frac{10}{3}c+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(c-\frac{5}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{5}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$c-\frac{5}{3}=\frac{2}{3} c-\frac{5}{3}=-\frac{2}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$c=\frac{7}{3} c=1$$