Solve 2x^2-36=x

$$2 x ^ { 2 } - 36 = x$$

Solve for x

$$x=-4$$

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$2x^{2}-36-x=0$$

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

$$2x^{2}-x-36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-36. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-36\right)=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=-9 b=8$$

Rewrite 2x^{2}-x-36 as \left(2x^{2}-9x\right)+\left(8x-36\right).

$$\left(2x^{2}-9x\right)+\left(8x-36\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(2x-9\right)+4\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x+4\right)$$

To find equation solutions, solve 2x-9=0 and x+4=0.

$$x=\frac{9}{2} x=-4$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$2x^{2}-36-x=0$$

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.

$$2x^{2}-x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-36\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-36\right)}}{2\times 2}$$

Multiply -8 times -36.

$$x=\frac{-\left(-1\right)±\sqrt{1+288}}{2\times 2}$$

Add 1 to 288.

$$x=\frac{-\left(-1\right)±\sqrt{289}}{2\times 2}$$

Take the square root of 289.

$$x=\frac{-\left(-1\right)±17}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±17}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±17}{4}$$

Now solve the equation x=\frac{1±17}{4} when ± is plus. Add 1 to 17.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{1±17}{4} when ± is minus. Subtract 17 from 1.

$$x=-\frac{16}{4}$$

Divide -16 by 4.

$$x=-4$$

The equation is now solved.

$$x=\frac{9}{2} x=-4$$

Steps for Completing the Square

Subtract x from both sides.

$$2x^{2}-36-x=0$$

Add 36 to both sides. Anything plus zero gives itself.

$$2x^{2}-x=36$$

Divide both sides by 2.

$$\frac{2x^{2}-x}{2}=\frac{36}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{1}{2}x=\frac{36}{2}$$

Divide 36 by 2.

$$x^{2}-\frac{1}{2}x=18$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=18+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=18+\frac{1}{16}$$

Add 18 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{289}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{289}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{289}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{17}{4} x-\frac{1}{4}=-\frac{17}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{9}{2} x=-4$$