Solve x^2-10/3x+x-3/2=1

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

Solve for x

$$x=-1$$

$$x=4$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 3x,2.

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

Use the distributive property to multiply 2 by x^{2}-10.

$$2x^{2}-20+3x\left(x-3\right)=6x$$

Use the distributive property to multiply 3x by x-3.

$$2x^{2}-20+3x^{2}-9x=6x$$

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

$$5x^{2}-20-9x=6x$$

Subtract 6x from both sides.

$$5x^{2}-20-9x-6x=0$$

Combine -9x and -6x to get -15x.

$$5x^{2}-20-15x=0$$

Divide both sides by 5.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

Rewrite x^{2}-3x-4 as \left(x^{2}-4x\right)+\left(x-4\right).

$$\left(x^{2}-4x\right)+\left(x-4\right)$$

Factor out x in x^{2}-4x.

$$x\left(x-4\right)+x-4$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+1\right)$$

To find equation solutions, solve x-4=0 and x+1=0.

$$x=4 x=-1$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 3x,2.

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

Use the distributive property to multiply 2 by x^{2}-10.

$$2x^{2}-20+3x\left(x-3\right)=6x$$

Use the distributive property to multiply 3x by x-3.

$$2x^{2}-20+3x^{2}-9x=6x$$

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

$$5x^{2}-20-9x=6x$$

Subtract 6x from both sides.

$$5x^{2}-20-9x-6x=0$$

Combine -9x and -6x to get -15x.

$$5x^{2}-20-15x=0$$

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.

$$5x^{2}-15x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -15 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 5\left(-20\right)}}{2\times 5}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 5\left(-20\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-15\right)±\sqrt{225-20\left(-20\right)}}{2\times 5}$$

Multiply -20 times -20.

$$x=\frac{-\left(-15\right)±\sqrt{225+400}}{2\times 5}$$

Add 225 to 400.

$$x=\frac{-\left(-15\right)±\sqrt{625}}{2\times 5}$$

Take the square root of 625.

$$x=\frac{-\left(-15\right)±25}{2\times 5}$$

The opposite of -15 is 15.

$$x=\frac{15±25}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{15±25}{10}$$

Now solve the equation x=\frac{15±25}{10} when ± is plus. Add 15 to 25.

$$x=\frac{40}{10}$$

Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{15±25}{10} when ± is minus. Subtract 25 from 15.

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

Divide -10 by 10.

$$x=-1$$

The equation is now solved.

$$x=4 x=-1$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6x, the least common multiple of 3x,2.

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

Use the distributive property to multiply 2 by x^{2}-10.

$$2x^{2}-20+3x\left(x-3\right)=6x$$

Use the distributive property to multiply 3x by x-3.

$$2x^{2}-20+3x^{2}-9x=6x$$

Combine 2x^{2} and 3x^{2} to get 5x^{2}.

$$5x^{2}-20-9x=6x$$

Subtract 6x from both sides.

$$5x^{2}-20-9x-6x=0$$

Combine -9x and -6x to get -15x.

$$5x^{2}-20-15x=0$$

Add 20 to both sides. Anything plus zero gives itself.

$$5x^{2}-15x=20$$

Divide both sides by 5.

$$\frac{5x^{2}-15x}{5}=\frac{20}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{15}{5}\right)x=\frac{20}{5}$$

Divide -15 by 5.

$$x^{2}-3x=\frac{20}{5}$$

Divide 20 by 5.

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

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}$$

Add 4 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{25}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=4 x=-1$$