Solve 3x^2-x/5=8

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

Solve for x

$$x = -\frac{8}{5} = -1\frac{3}{5} = -1.6$$

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

Steps Using Factoring By Grouping

Multiply both sides of the equation by 5.

$$15x^{2}-x=40$$

Subtract 40 from both sides.

$$15x^{2}-x-40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-40. To find a and b, set up a system to be solved.

$$a+b=-1 ab=15\left(-40\right)=-600$$

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 -600.

$$1,-600 2,-300 3,-200 4,-150 5,-120 6,-100 8,-75 10,-60 12,-50 15,-40 20,-30 24,-25$$

Calculate the sum for each pair.

$$1-600=-599 2-300=-298 3-200=-197 4-150=-146 5-120=-115 6-100=-94 8-75=-67 10-60=-50 12-50=-38 15-40=-25 20-30=-10 24-25=-1$$

The solution is the pair that gives sum -1.

$$a=-25 b=24$$

Rewrite 15x^{2}-x-40 as \left(15x^{2}-25x\right)+\left(24x-40\right).

$$\left(15x^{2}-25x\right)+\left(24x-40\right)$$

Factor out 5x in the first and 8 in the second group.

$$5x\left(3x-5\right)+8\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(5x+8\right)$$

To find equation solutions, solve 3x-5=0 and 5x+8=0.

$$x=\frac{5}{3} x=-\frac{8}{5}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$15x^{2}-x=40$$

Subtract 40 from both sides.

$$15x^{2}-x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -1 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 15\left(-40\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-1\right)±\sqrt{1-60\left(-40\right)}}{2\times 15}$$

Multiply -60 times -40.

$$x=\frac{-\left(-1\right)±\sqrt{1+2400}}{2\times 15}$$

Add 1 to 2400.

$$x=\frac{-\left(-1\right)±\sqrt{2401}}{2\times 15}$$

Take the square root of 2401.

$$x=\frac{-\left(-1\right)±49}{2\times 15}$$

The opposite of -1 is 1.

$$x=\frac{1±49}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{1±49}{30}$$

Now solve the equation x=\frac{1±49}{30} when ± is plus. Add 1 to 49.

$$x=\frac{50}{30}$$

Reduce the fraction \frac{50}{30} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{1±49}{30} when ± is minus. Subtract 49 from 1.

$$x=-\frac{48}{30}$$

Reduce the fraction \frac{-48}{30} to lowest terms by extracting and canceling out 6.

$$x=-\frac{8}{5}$$

The equation is now solved.

$$x=\frac{5}{3} x=-\frac{8}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$15x^{2}-x=40$$

Divide both sides by 15.

$$\frac{15x^{2}-x}{15}=\frac{40}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}-\frac{1}{15}x=\frac{40}{15}$$

Reduce the fraction \frac{40}{15} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{1}{15}x=\frac{8}{3}$$

Divide -\frac{1}{15}, the coefficient of the x term, by 2 to get -\frac{1}{30}. Then add the square of -\frac{1}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{15}x+\left(-\frac{1}{30}\right)^{2}=\frac{8}{3}+\left(-\frac{1}{30}\right)^{2}$$

Square -\frac{1}{30} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{15}x+\frac{1}{900}=\frac{8}{3}+\frac{1}{900}$$

Add \frac{8}{3} to \frac{1}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{15}x+\frac{1}{900}=\frac{2401}{900}$$

Factor x^{2}-\frac{1}{15}x+\frac{1}{900}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{30}\right)^{2}=\frac{2401}{900}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{30}\right)^{2}}=\sqrt{\frac{2401}{900}}$$

Simplify.

$$x-\frac{1}{30}=\frac{49}{30} x-\frac{1}{30}=-\frac{49}{30}$$

Add \frac{1}{30} to both sides of the equation.

$$x=\frac{5}{3} x=-\frac{8}{5}$$