Solve 3x^2-7x-3=x

$$3 x ^ { 2 } - 7 x - 3 = x$$

Solve for x

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

$$x=3$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$3x^{2}-7x-3-x=0$$

Combine -7x and -x to get -8x.

$$3x^{2}-8x-3=0$$

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

$$a+b=-8 ab=3\left(-3\right)=-9$$

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. List all such integer pairs that give product -9.

$$1,-9 3,-3$$

Calculate the sum for each pair.

$$1-9=-8 3-3=0$$

The solution is the pair that gives sum -8.

$$a=-9 b=1$$

Rewrite 3x^{2}-8x-3 as \left(3x^{2}-9x\right)+\left(x-3\right).

$$\left(3x^{2}-9x\right)+\left(x-3\right)$$

Factor out 3x in 3x^{2}-9x.

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

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

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

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

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

Steps Using the Quadratic Formula

Subtract x from both sides.

$$3x^{2}-7x-3-x=0$$

Combine -7x and -x to get -8x.

$$3x^{2}-8x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\left(-3\right)}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\left(-3\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\left(-3\right)}}{2\times 3}$$

Multiply -12 times -3.

$$x=\frac{-\left(-8\right)±\sqrt{64+36}}{2\times 3}$$

Add 64 to 36.

$$x=\frac{-\left(-8\right)±\sqrt{100}}{2\times 3}$$

Take the square root of 100.

$$x=\frac{-\left(-8\right)±10}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±10}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±10}{6}$$

Now solve the equation x=\frac{8±10}{6} when ± is plus. Add 8 to 10.

$$x=\frac{18}{6}$$

Divide 18 by 6.

$$x=3$$

Now solve the equation x=\frac{8±10}{6} when ± is minus. Subtract 10 from 8.

$$x=-\frac{2}{6}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

$$3x^{2}-7x-3-x=0$$

Combine -7x and -x to get -8x.

$$3x^{2}-8x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$3x^{2}-8x=3$$

Divide both sides by 3.

$$\frac{3x^{2}-8x}{3}=\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{8}{3}x=\frac{3}{3}$$

Divide 3 by 3.

$$x^{2}-\frac{8}{3}x=1$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=1+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=1+\frac{16}{9}$$

Add 1 to \frac{16}{9}.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{25}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{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}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{5}{3} x-\frac{4}{3}=-\frac{5}{3}$$

Add \frac{4}{3} to both sides of the equation.

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