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

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

Solve for x

$$x = -\frac{14}{5} = -2\frac{4}{5} = -2.8$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right).

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

Use the distributive property to multiply 5 by x-2.

$$x-2=\left(5x-10\right)\left(x+3\right)$$

Use the distributive property to multiply 5x-10 by x+3 and combine like terms.

$$x-2=5x^{2}+5x-30$$

Subtract 5x^{2} from both sides.

$$x-2-5x^{2}=5x-30$$

Subtract 5x from both sides.

$$x-2-5x^{2}-5x=-30$$

Combine x and -5x to get -4x.

$$-4x-2-5x^{2}=-30$$

Add 30 to both sides.

$$-4x-2-5x^{2}+30=0$$

Add -2 and 30 to get 28.

$$-4x+28-5x^{2}=0$$

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

$$-5x^{2}-4x+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx+28. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-5\times 28=-140$$

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 -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -4.

$$a=10 b=-14$$

Rewrite -5x^{2}-4x+28 as \left(-5x^{2}+10x\right)+\left(-14x+28\right).

$$\left(-5x^{2}+10x\right)+\left(-14x+28\right)$$

Factor out 5x in the first and 14 in the second group.

$$5x\left(-x+2\right)+14\left(-x+2\right)$$

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

$$\left(-x+2\right)\left(5x+14\right)$$

To find equation solutions, solve -x+2=0 and 5x+14=0.

$$x=2 x=-\frac{14}{5}$$

Variable x cannot be equal to 2.

$$x=-\frac{14}{5}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right).

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

Use the distributive property to multiply 5 by x-2.

$$x-2=\left(5x-10\right)\left(x+3\right)$$

Use the distributive property to multiply 5x-10 by x+3 and combine like terms.

$$x-2=5x^{2}+5x-30$$

Subtract 5x^{2} from both sides.

$$x-2-5x^{2}=5x-30$$

Subtract 5x from both sides.

$$x-2-5x^{2}-5x=-30$$

Combine x and -5x to get -4x.

$$-4x-2-5x^{2}=-30$$

Add 30 to both sides.

$$-4x-2-5x^{2}+30=0$$

Add -2 and 30 to get 28.

$$-4x+28-5x^{2}=0$$

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.

$$-5x^{2}-4x+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -4 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-5\right)\times 28}}{2\left(-5\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-5\right)\times 28}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-4\right)±\sqrt{16+20\times 28}}{2\left(-5\right)}$$

Multiply 20 times 28.

$$x=\frac{-\left(-4\right)±\sqrt{16+560}}{2\left(-5\right)}$$

Add 16 to 560.

$$x=\frac{-\left(-4\right)±\sqrt{576}}{2\left(-5\right)}$$

Take the square root of 576.

$$x=\frac{-\left(-4\right)±24}{2\left(-5\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±24}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{4±24}{-10}$$

Now solve the equation x=\frac{4±24}{-10} when ± is plus. Add 4 to 24.

$$x=\frac{28}{-10}$$

Reduce the fraction \frac{28}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{14}{5}$$

Now solve the equation x=\frac{4±24}{-10} when ± is minus. Subtract 24 from 4.

$$x=-\frac{20}{-10}$$

Divide -20 by -10.

$$x=2$$

The equation is now solved.

$$x=-\frac{14}{5} x=2$$

Variable x cannot be equal to 2.

$$x=-\frac{14}{5}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+3\right).

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

Use the distributive property to multiply 5 by x-2.

$$x-2=\left(5x-10\right)\left(x+3\right)$$

Use the distributive property to multiply 5x-10 by x+3 and combine like terms.

$$x-2=5x^{2}+5x-30$$

Subtract 5x^{2} from both sides.

$$x-2-5x^{2}=5x-30$$

Subtract 5x from both sides.

$$x-2-5x^{2}-5x=-30$$

Combine x and -5x to get -4x.

$$-4x-2-5x^{2}=-30$$

Add 2 to both sides.

$$-4x-5x^{2}=-30+2$$

Add -30 and 2 to get -28.

$$-4x-5x^{2}=-28$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-5x^{2}-4x=-28$$

Divide both sides by -5.

$$\frac{-5x^{2}-4x}{-5}=-\frac{28}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{4}{-5}\right)x=-\frac{28}{-5}$$

Divide -4 by -5.

$$x^{2}+\frac{4}{5}x=-\frac{28}{-5}$$

Divide -28 by -5.

$$x^{2}+\frac{4}{5}x=\frac{28}{5}$$

Divide \frac{4}{5}, the coefficient of the x term, by 2 to get \frac{2}{5}. Then add the square of \frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\frac{28}{5}+\left(\frac{2}{5}\right)^{2}$$

Square \frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{28}{5}+\frac{4}{25}$$

Add \frac{28}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{144}{25}$$

Factor x^{2}+\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{144}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{144}{25}}$$

Simplify.

$$x+\frac{2}{5}=\frac{12}{5} x+\frac{2}{5}=-\frac{12}{5}$$

Subtract \frac{2}{5} from both sides of the equation.

$$x=2 x=-\frac{14}{5}$$

Variable x cannot be equal to 2.

$$x=-\frac{14}{5}$$