Solve 2x+12/4=(x(x))

$$\frac{ 2x+12 }{ 4 } = (x(x))$$

Solve for x

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

$$x=2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4.

$$2x+12=4xx$$

Multiply x and x to get x^{2}.

$$2x+12=4x^{2}$$

Subtract 4x^{2} from both sides.

$$2x+12-4x^{2}=0$$

Divide both sides by 2.

$$x+6-2x^{2}=0$$

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

$$-2x^{2}+x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=1 ab=-2\times 6=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

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

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -2x^{2}+x+6 as \left(-2x^{2}+4x\right)+\left(-3x+6\right).

$$\left(-2x^{2}+4x\right)+\left(-3x+6\right)$$

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

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

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

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

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

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

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4.

$$2x+12=4xx$$

Multiply x and x to get x^{2}.

$$2x+12=4x^{2}$$

Subtract 4x^{2} from both sides.

$$2x+12-4x^{2}=0$$

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.

$$-4x^{2}+2x+12=0$$

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

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

Square 2.

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

Multiply -4 times -4.

$$x=\frac{-2±\sqrt{4+16\times 12}}{2\left(-4\right)}$$

Multiply 16 times 12.

$$x=\frac{-2±\sqrt{4+192}}{2\left(-4\right)}$$

Add 4 to 192.

$$x=\frac{-2±\sqrt{196}}{2\left(-4\right)}$$

Take the square root of 196.

$$x=\frac{-2±14}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±14}{-8}$$

Now solve the equation x=\frac{-2±14}{-8} when ± is plus. Add -2 to 14.

$$x=\frac{12}{-8}$$

Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-2±14}{-8} when ± is minus. Subtract 14 from -2.

$$x=-\frac{16}{-8}$$

Divide -16 by -8.

$$x=2$$

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 4.

$$2x+12=4xx$$

Multiply x and x to get x^{2}.

$$2x+12=4x^{2}$$

Subtract 4x^{2} from both sides.

$$2x+12-4x^{2}=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$2x-4x^{2}=-12$$

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.

$$-4x^{2}+2x=-12$$

Divide both sides by -4.

$$\frac{-4x^{2}+2x}{-4}=-\frac{12}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{2}{-4}x=-\frac{12}{-4}$$

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

$$x^{2}-\frac{1}{2}x=-\frac{12}{-4}$$

Divide -12 by -4.

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

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=3+\left(-\frac{1}{4}\right)^{2}$$

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=3+\frac{1}{16}$$

Add 3 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{49}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{7}{4} x-\frac{1}{4}=-\frac{7}{4}$$

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

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