Solve 5x^2+x-1=1/3(x-2)

$$5 x ^ { 2 } + x - 1 = \frac { 1 } { 3 } ( x - 2 )$$

Solve for x

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

$$x=\frac{1}{5}=0.2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{3} by x-2.

$$5x^{2}+x-1=\frac{1}{3}x-\frac{2}{3}$$

Subtract \frac{1}{3}x from both sides.

$$5x^{2}+x-1-\frac{1}{3}x=-\frac{2}{3}$$

Combine x and -\frac{1}{3}x to get \frac{2}{3}x.

$$5x^{2}+\frac{2}{3}x-1=-\frac{2}{3}$$

Add \frac{2}{3} to both sides.

$$5x^{2}+\frac{2}{3}x-1+\frac{2}{3}=0$$

Add -1 and \frac{2}{3} to get -\frac{1}{3}.

$$5x^{2}+\frac{2}{3}x-\frac{1}{3}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, \frac{2}{3} for b, and -\frac{1}{3} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{2}{3}±\sqrt{\left(\frac{2}{3}\right)^{2}-4\times 5\left(-\frac{1}{3}\right)}}{2\times 5}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-4\times 5\left(-\frac{1}{3}\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}-20\left(-\frac{1}{3}\right)}}{2\times 5}$$

Multiply -20 times -\frac{1}{3}.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{4}{9}+\frac{20}{3}}}{2\times 5}$$

Add \frac{4}{9} to \frac{20}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{2}{3}±\sqrt{\frac{64}{9}}}{2\times 5}$$

Take the square root of \frac{64}{9}.

$$x=\frac{-\frac{2}{3}±\frac{8}{3}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-\frac{2}{3}±\frac{8}{3}}{10}$$

Now solve the equation x=\frac{-\frac{2}{3}±\frac{8}{3}}{10} when ± is plus. Add -\frac{2}{3} to \frac{8}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{2}{10}$$

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

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-\frac{2}{3}±\frac{8}{3}}{10} when ± is minus. Subtract \frac{8}{3} from -\frac{2}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{10}{3}}{10}$$

Divide -\frac{10}{3} by 10.

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

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{1}{3}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{3} by x-2.

$$5x^{2}+x-1=\frac{1}{3}x-\frac{2}{3}$$

Subtract \frac{1}{3}x from both sides.

$$5x^{2}+x-1-\frac{1}{3}x=-\frac{2}{3}$$

Combine x and -\frac{1}{3}x to get \frac{2}{3}x.

$$5x^{2}+\frac{2}{3}x-1=-\frac{2}{3}$$

Add 1 to both sides.

$$5x^{2}+\frac{2}{3}x=-\frac{2}{3}+1$$

Add -\frac{2}{3} and 1 to get \frac{1}{3}.

$$5x^{2}+\frac{2}{3}x=\frac{1}{3}$$

Divide both sides by 5.

$$\frac{5x^{2}+\frac{2}{3}x}{5}=\frac{\frac{1}{3}}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{\frac{2}{3}}{5}x=\frac{\frac{1}{3}}{5}$$

Divide \frac{2}{3} by 5.

$$x^{2}+\frac{2}{15}x=\frac{\frac{1}{3}}{5}$$

Divide \frac{1}{3} by 5.

$$x^{2}+\frac{2}{15}x=\frac{1}{15}$$

Divide \frac{2}{15}, the coefficient of the x term, by 2 to get \frac{1}{15}. Then add the square of \frac{1}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{15}x+\left(\frac{1}{15}\right)^{2}=\frac{1}{15}+\left(\frac{1}{15}\right)^{2}$$

Square \frac{1}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{15}x+\frac{1}{225}=\frac{1}{15}+\frac{1}{225}$$

Add \frac{1}{15} to \frac{1}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{15}x+\frac{1}{225}=\frac{16}{225}$$

Factor x^{2}+\frac{2}{15}x+\frac{1}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{15}\right)^{2}=\frac{16}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{15}\right)^{2}}=\sqrt{\frac{16}{225}}$$

Simplify.

$$x+\frac{1}{15}=\frac{4}{15} x+\frac{1}{15}=-\frac{4}{15}$$

Subtract \frac{1}{15} from both sides of the equation.

$$x=\frac{1}{5} x=-\frac{1}{3}$$