Solve 2x^2-12=5x

$$2 x ^ { 2 } - 12 = 5 x$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=4$$

Steps Using Factoring By Grouping

Subtract 5x from both sides.

$$2x^{2}-12-5x=0$$

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

$$2x^{2}-5x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\left(-12\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -5.

$$a=-8 b=3$$

Rewrite 2x^{2}-5x-12 as \left(2x^{2}-8x\right)+\left(3x-12\right).

$$\left(2x^{2}-8x\right)+\left(3x-12\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(x-4\right)+3\left(x-4\right)$$

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

$$\left(x-4\right)\left(2x+3\right)$$

To find equation solutions, solve x-4=0 and 2x+3=0.

$$x=4 x=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Subtract 5x from both sides.

$$2x^{2}-12-5x=0$$

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.

$$2x^{2}-5x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-12\right)}}{2\times 2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-12\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-5\right)±\sqrt{25-8\left(-12\right)}}{2\times 2}$$

Multiply -8 times -12.

$$x=\frac{-\left(-5\right)±\sqrt{25+96}}{2\times 2}$$

Add 25 to 96.

$$x=\frac{-\left(-5\right)±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-\left(-5\right)±11}{2\times 2}$$

The opposite of -5 is 5.

$$x=\frac{5±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{5±11}{4}$$

Now solve the equation x=\frac{5±11}{4} when ± is plus. Add 5 to 11.

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

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{5±11}{4} when ± is minus. Subtract 11 from 5.

$$x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

$$x=4 x=-\frac{3}{2}$$

Steps for Completing the Square

Subtract 5x from both sides.

$$2x^{2}-12-5x=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$2x^{2}-5x=12$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 12 by 2.

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

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

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=6+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=6+\frac{25}{16}$$

Add 6 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{121}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{11}{4} x-\frac{5}{4}=-\frac{11}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=4 x=-\frac{3}{2}$$