Solve 14-3xx=-x+4

$$14 - 3 x x = - x + 4$$

Solve for x

$$x = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$x=2$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$14-3x^{2}=-x+4$$

Add x to both sides.

$$14-3x^{2}+x=4$$

Subtract 4 from both sides.

$$14-3x^{2}+x-4=0$$

Subtract 4 from 14 to get 10.

$$10-3x^{2}+x=0$$

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.

$$-3x^{2}+x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 1 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-3\right)\times 10}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-1±\sqrt{1+12\times 10}}{2\left(-3\right)}$$

Multiply 12 times 10.

$$x=\frac{-1±\sqrt{1+120}}{2\left(-3\right)}$$

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2\left(-3\right)}$$

Take the square root of 121.

$$x=\frac{-1±11}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-1±11}{-6}$$

Now solve the equation x=\frac{-1±11}{-6} when ± is plus. Add -1 to 11.

$$x=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-1±11}{-6} when ± is minus. Subtract 11 from -1.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$14-3x^{2}=-x+4$$

Add x to both sides.

$$14-3x^{2}+x=4$$

Subtract 14 from both sides.

$$-3x^{2}+x=4-14$$

Subtract 14 from 4 to get -10.

$$-3x^{2}+x=-10$$

Divide both sides by -3.

$$\frac{-3x^{2}+x}{-3}=-\frac{10}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{1}{-3}x=-\frac{10}{-3}$$

Divide 1 by -3.

$$x^{2}-\frac{1}{3}x=-\frac{10}{-3}$$

Divide -10 by -3.

$$x^{2}-\frac{1}{3}x=\frac{10}{3}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{10}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{10}{3}+\frac{1}{36}$$

Add \frac{10}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{121}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{11}{6} x-\frac{1}{6}=-\frac{11}{6}$$

Add \frac{1}{6} to both sides of the equation.

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