Solve 21c^2-16c-6=1-2c

$$21 c ^ { 2 } - 16 c - 6 = 1 - 2 c$$

Solve for c

$$c=-\frac{1}{3}\approx -0.333333333$$

$$c=1$$

Steps Using Factoring By Grouping

Subtract 1 from both sides.

$$21c^{2}-16c-6-1=-2c$$

Subtract 1 from -6 to get -7.

$$21c^{2}-16c-7=-2c$$

Add 2c to both sides.

$$21c^{2}-16c-7+2c=0$$

Combine -16c and 2c to get -14c.

$$21c^{2}-14c-7=0$$

Divide both sides by 7.

$$3c^{2}-2c-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3c^{2}+ac+bc-1. To find a and b, set up a system to be solved.

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

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 3c^{2}-2c-1 as \left(3c^{2}-3c\right)+\left(c-1\right).

$$\left(3c^{2}-3c\right)+\left(c-1\right)$$

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

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

Factor out common term c-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$21c^{2}-16c-6-1=-2c$$

Subtract 1 from -6 to get -7.

$$21c^{2}-16c-7=-2c$$

Add 2c to both sides.

$$21c^{2}-16c-7+2c=0$$

Combine -16c and 2c to get -14c.

$$21c^{2}-14c-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -14 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 21\left(-7\right)}}{2\times 21}$$

Square -14.

$$c=\frac{-\left(-14\right)±\sqrt{196-4\times 21\left(-7\right)}}{2\times 21}$$

Multiply -4 times 21.

$$c=\frac{-\left(-14\right)±\sqrt{196-84\left(-7\right)}}{2\times 21}$$

Multiply -84 times -7.

$$c=\frac{-\left(-14\right)±\sqrt{196+588}}{2\times 21}$$

Add 196 to 588.

$$c=\frac{-\left(-14\right)±\sqrt{784}}{2\times 21}$$

Take the square root of 784.

$$c=\frac{-\left(-14\right)±28}{2\times 21}$$

The opposite of -14 is 14.

$$c=\frac{14±28}{2\times 21}$$

Multiply 2 times 21.

$$c=\frac{14±28}{42}$$

Now solve the equation c=\frac{14±28}{42} when ± is plus. Add 14 to 28.

$$c=\frac{42}{42}$$

Divide 42 by 42.

$$c=1$$

Now solve the equation c=\frac{14±28}{42} when ± is minus. Subtract 28 from 14.

$$c=-\frac{14}{42}$$

Reduce the fraction \frac{-14}{42} to lowest terms by extracting and canceling out 14.

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

The equation is now solved.

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

Steps for Completing the Square

Add 2c to both sides.

$$21c^{2}-16c-6+2c=1$$

Combine -16c and 2c to get -14c.

$$21c^{2}-14c-6=1$$

Add 6 to both sides.

$$21c^{2}-14c=1+6$$

Add 1 and 6 to get 7.

$$21c^{2}-14c=7$$

Divide both sides by 21.

$$\frac{21c^{2}-14c}{21}=\frac{7}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$c^{2}+\left(-\frac{14}{21}\right)c=\frac{7}{21}$$

Reduce the fraction \frac{-14}{21} to lowest terms by extracting and canceling out 7.

$$c^{2}-\frac{2}{3}c=\frac{7}{21}$$

Reduce the fraction \frac{7}{21} to lowest terms by extracting and canceling out 7.

$$c^{2}-\frac{2}{3}c=\frac{1}{3}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$c^{2}-\frac{2}{3}c+\left(-\frac{1}{3}\right)^{2}=\frac{1}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$c^{2}-\frac{2}{3}c+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}$$

Add \frac{1}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$c^{2}-\frac{2}{3}c+\frac{1}{9}=\frac{4}{9}$$

Factor c^{2}-\frac{2}{3}c+\frac{1}{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}.

$$\left(c-\frac{1}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(c-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

$$c-\frac{1}{3}=\frac{2}{3} c-\frac{1}{3}=-\frac{2}{3}$$

Add \frac{1}{3} to both sides of the equation.

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