Solve 2x^2-10/x+5-4=0

$$\frac{ 2 { x }^{ 2 } -10 }{ x+5 } -4=0$$

Solve for x

$$x=-3$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -5 since division by zero is not defined. Multiply both sides of the equation by x+5.

$$2x^{2}-10+\left(x+5\right)\left(-4\right)=0$$

Use the distributive property to multiply x+5 by -4.

$$2x^{2}-10-4x-20=0$$

Subtract 20 from -10 to get -30.

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

Divide both sides by 2.

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

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

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

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

$$a+b=-2 ab=1\left(-15\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite x^{2}-2x-15 as \left(x^{2}-5x\right)+\left(3x-15\right).

$$\left(x^{2}-5x\right)+\left(3x-15\right)$$

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

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

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

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

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

$$x=5 x=-3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -5 since division by zero is not defined. Multiply both sides of the equation by x+5.

$$2x^{2}-10+\left(x+5\right)\left(-4\right)=0$$

Use the distributive property to multiply x+5 by -4.

$$2x^{2}-10-4x-20=0$$

Subtract 20 from -10 to get -30.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-30\right)}}{2\times 2}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-30\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-8\left(-30\right)}}{2\times 2}$$

Multiply -8 times -30.

$$x=\frac{-\left(-4\right)±\sqrt{16+240}}{2\times 2}$$

Add 16 to 240.

$$x=\frac{-\left(-4\right)±\sqrt{256}}{2\times 2}$$

Take the square root of 256.

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

The opposite of -4 is 4.

$$x=\frac{4±16}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{4±16}{4}$$

Now solve the equation x=\frac{4±16}{4} when ± is plus. Add 4 to 16.

$$x=\frac{20}{4}$$

Divide 20 by 4.

$$x=5$$

Now solve the equation x=\frac{4±16}{4} when ± is minus. Subtract 16 from 4.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=5 x=-3$$

Steps for Completing the Square

Variable x cannot be equal to -5 since division by zero is not defined. Multiply both sides of the equation by x+5.

$$2x^{2}-10+\left(x+5\right)\left(-4\right)=0$$

Use the distributive property to multiply x+5 by -4.

$$2x^{2}-10-4x-20=0$$

Subtract 20 from -10 to get -30.

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

Add 30 to both sides. Anything plus zero gives itself.

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

Divide both sides by 2.

$$\frac{2x^{2}-4x}{2}=\frac{30}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{4}{2}\right)x=\frac{30}{2}$$

Divide -4 by 2.

$$x^{2}-2x=\frac{30}{2}$$

Divide 30 by 2.

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

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

$$x^{2}-2x+1=15+1$$

Add 15 to 1.

$$x^{2}-2x+1=16$$

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

$$\left(x-1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-1=4 x-1=-4$$

Add 1 to both sides of the equation.

$$x=5 x=-3$$