Solve x^2/5-2x/3=x-5/6

$$\frac{ { x }^{ 2 } }{ 5 } - \frac{ 2x }{ 3 } = \frac{ x-5 }{ 6 }$$

Solve for x

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

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 30, the least common multiple of 5,3,6.

$$6x^{2}-10\times 2x=5\left(x-5\right)$$

Multiply -10 and 2 to get -20.

$$6x^{2}-20x=5\left(x-5\right)$$

Use the distributive property to multiply 5 by x-5.

$$6x^{2}-20x=5x-25$$

Subtract 5x from both sides.

$$6x^{2}-20x-5x=-25$$

Combine -20x and -5x to get -25x.

$$6x^{2}-25x=-25$$

Add 25 to both sides.

$$6x^{2}-25x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=-25 ab=6\times 25=150$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 150.

$$-1,-150 -2,-75 -3,-50 -5,-30 -6,-25 -10,-15$$

Calculate the sum for each pair.

$$-1-150=-151 -2-75=-77 -3-50=-53 -5-30=-35 -6-25=-31 -10-15=-25$$

The solution is the pair that gives sum -25.

$$a=-15 b=-10$$

Rewrite 6x^{2}-25x+25 as \left(6x^{2}-15x\right)+\left(-10x+25\right).

$$\left(6x^{2}-15x\right)+\left(-10x+25\right)$$

Factor out 3x in the first and -5 in the second group.

$$3x\left(2x-5\right)-5\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(3x-5\right)$$

To find equation solutions, solve 2x-5=0 and 3x-5=0.

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 30, the least common multiple of 5,3,6.

$$6x^{2}-10\times 2x=5\left(x-5\right)$$

Multiply -10 and 2 to get -20.

$$6x^{2}-20x=5\left(x-5\right)$$

Use the distributive property to multiply 5 by x-5.

$$6x^{2}-20x=5x-25$$

Subtract 5x from both sides.

$$6x^{2}-20x-5x=-25$$

Combine -20x and -5x to get -25x.

$$6x^{2}-25x=-25$$

Add 25 to both sides.

$$6x^{2}-25x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -25 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 6\times 25}}{2\times 6}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 6\times 25}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-25\right)±\sqrt{625-24\times 25}}{2\times 6}$$

Multiply -24 times 25.

$$x=\frac{-\left(-25\right)±\sqrt{625-600}}{2\times 6}$$

Add 625 to -600.

$$x=\frac{-\left(-25\right)±\sqrt{25}}{2\times 6}$$

Take the square root of 25.

$$x=\frac{-\left(-25\right)±5}{2\times 6}$$

The opposite of -25 is 25.

$$x=\frac{25±5}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{25±5}{12}$$

Now solve the equation x=\frac{25±5}{12} when ± is plus. Add 25 to 5.

$$x=\frac{30}{12}$$

Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{25±5}{12} when ± is minus. Subtract 5 from 25.

$$x=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 30, the least common multiple of 5,3,6.

$$6x^{2}-10\times 2x=5\left(x-5\right)$$

Multiply -10 and 2 to get -20.

$$6x^{2}-20x=5\left(x-5\right)$$

Use the distributive property to multiply 5 by x-5.

$$6x^{2}-20x=5x-25$$

Subtract 5x from both sides.

$$6x^{2}-20x-5x=-25$$

Combine -20x and -5x to get -25x.

$$6x^{2}-25x=-25$$

Divide both sides by 6.

$$\frac{6x^{2}-25x}{6}=-\frac{25}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{25}{6}x=-\frac{25}{6}$$

Divide -\frac{25}{6}, the coefficient of the x term, by 2 to get -\frac{25}{12}. Then add the square of -\frac{25}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{25}{6}x+\left(-\frac{25}{12}\right)^{2}=-\frac{25}{6}+\left(-\frac{25}{12}\right)^{2}$$

Square -\frac{25}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{25}{6}x+\frac{625}{144}=-\frac{25}{6}+\frac{625}{144}$$

Add -\frac{25}{6} to \frac{625}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{25}{6}x+\frac{625}{144}=\frac{25}{144}$$

Factor x^{2}-\frac{25}{6}x+\frac{625}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{25}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$x-\frac{25}{12}=\frac{5}{12} x-\frac{25}{12}=-\frac{5}{12}$$

Add \frac{25}{12} to both sides of the equation.

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