Solve -1=3x^2-4(x+2)+x

$$- 1 = 3 x ^ { 2 } - 4 ( x + 2 ) + x$$

Solve for x

$$x=\frac{\sqrt{93}}{6}+\frac{1}{2}\approx 2.107275127$$

$$x=-\frac{\sqrt{93}}{6}+\frac{1}{2}\approx -1.107275127$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -4 by x+2.

$$-1=3x^{2}-4x-8+x$$

Combine -4x and x to get -3x.

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

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

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

Add 1 to both sides.

$$3x^{2}-3x-8+1=0$$

Add -8 and 1 to get -7.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-7\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-3\right)±\sqrt{9-12\left(-7\right)}}{2\times 3}$$

Multiply -12 times -7.

$$x=\frac{-\left(-3\right)±\sqrt{9+84}}{2\times 3}$$

Add 9 to 84.

$$x=\frac{-\left(-3\right)±\sqrt{93}}{2\times 3}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{93}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{3±\sqrt{93}}{6}$$

Now solve the equation x=\frac{3±\sqrt{93}}{6} when ± is plus. Add 3 to \sqrt{93}.

$$x=\frac{\sqrt{93}+3}{6}$$

Divide 3+\sqrt{93} by 6.

$$x=\frac{\sqrt{93}}{6}+\frac{1}{2}$$

Now solve the equation x=\frac{3±\sqrt{93}}{6} when ± is minus. Subtract \sqrt{93} from 3.

$$x=\frac{3-\sqrt{93}}{6}$$

Divide 3-\sqrt{93} by 6.

$$x=-\frac{\sqrt{93}}{6}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{93}}{6}+\frac{1}{2} x=-\frac{\sqrt{93}}{6}+\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply -4 by x+2.

$$-1=3x^{2}-4x-8+x$$

Combine -4x and x to get -3x.

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

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

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

Add 8 to both sides.

$$3x^{2}-3x=-1+8$$

Add -1 and 8 to get 7.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

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

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{7}{3}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=\frac{7}{3}+\frac{1}{4}$$

Add \frac{7}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=\frac{31}{12}$$

Factor x^{2}-x+\frac{1}{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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{31}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{31}{12}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{93}}{6} x-\frac{1}{2}=-\frac{\sqrt{93}}{6}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{93}}{6}+\frac{1}{2} x=-\frac{\sqrt{93}}{6}+\frac{1}{2}$$