Solve 4x-9+12x=3/2x-2x

$$4x-9+12x= \frac{ 3 }{ 2x } -2x$$

Solve for x

$$x=\frac{\sqrt{21}}{12}+\frac{1}{4}\approx 0.631881308$$

$$x=-\frac{\sqrt{21}}{12}+\frac{1}{4}\approx -0.131881308$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$4x\times 2x+2x\left(-9\right)+12x\times 2x=3-2x\times 2x$$

Multiply 4 and 2 to get 8.

$$8xx+2x\left(-9\right)+12x\times 2x=3-2x\times 2x$$

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

$$8x^{2}+2x\left(-9\right)+12x\times 2x=3-2x\times 2x$$

Multiply 2 and -9 to get -18.

$$8x^{2}-18x+12x\times 2x=3-2x\times 2x$$

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

$$8x^{2}-18x+12x^{2}\times 2=3-2x\times 2x$$

Multiply 12 and 2 to get 24.

$$8x^{2}-18x+24x^{2}=3-2x\times 2x$$

Combine 8x^{2} and 24x^{2} to get 32x^{2}.

$$32x^{2}-18x=3-2x\times 2x$$

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

$$32x^{2}-18x=3-2x^{2}\times 2$$

Multiply -2 and 2 to get -4.

$$32x^{2}-18x=3-4x^{2}$$

Subtract 3 from both sides.

$$32x^{2}-18x-3=-4x^{2}$$

Add 4x^{2} to both sides.

$$32x^{2}-18x-3+4x^{2}=0$$

Combine 32x^{2} and 4x^{2} to get 36x^{2}.

$$36x^{2}-18x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, -18 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 36\left(-3\right)}}{2\times 36}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 36\left(-3\right)}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-\left(-18\right)±\sqrt{324-144\left(-3\right)}}{2\times 36}$$

Multiply -144 times -3.

$$x=\frac{-\left(-18\right)±\sqrt{324+432}}{2\times 36}$$

Add 324 to 432.

$$x=\frac{-\left(-18\right)±\sqrt{756}}{2\times 36}$$

Take the square root of 756.

$$x=\frac{-\left(-18\right)±6\sqrt{21}}{2\times 36}$$

The opposite of -18 is 18.

$$x=\frac{18±6\sqrt{21}}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{18±6\sqrt{21}}{72}$$

Now solve the equation x=\frac{18±6\sqrt{21}}{72} when ± is plus. Add 18 to 6\sqrt{21}.

$$x=\frac{6\sqrt{21}+18}{72}$$

Divide 18+6\sqrt{21} by 72.

$$x=\frac{\sqrt{21}}{12}+\frac{1}{4}$$

Now solve the equation x=\frac{18±6\sqrt{21}}{72} when ± is minus. Subtract 6\sqrt{21} from 18.

$$x=\frac{18-6\sqrt{21}}{72}$$

Divide 18-6\sqrt{21} by 72.

$$x=-\frac{\sqrt{21}}{12}+\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{21}}{12}+\frac{1}{4} x=-\frac{\sqrt{21}}{12}+\frac{1}{4}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 2x.

$$4x\times 2x+2x\left(-9\right)+12x\times 2x=3-2x\times 2x$$

Multiply 4 and 2 to get 8.

$$8xx+2x\left(-9\right)+12x\times 2x=3-2x\times 2x$$

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

$$8x^{2}+2x\left(-9\right)+12x\times 2x=3-2x\times 2x$$

Multiply 2 and -9 to get -18.

$$8x^{2}-18x+12x\times 2x=3-2x\times 2x$$

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

$$8x^{2}-18x+12x^{2}\times 2=3-2x\times 2x$$

Multiply 12 and 2 to get 24.

$$8x^{2}-18x+24x^{2}=3-2x\times 2x$$

Combine 8x^{2} and 24x^{2} to get 32x^{2}.

$$32x^{2}-18x=3-2x\times 2x$$

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

$$32x^{2}-18x=3-2x^{2}\times 2$$

Multiply -2 and 2 to get -4.

$$32x^{2}-18x=3-4x^{2}$$

Add 4x^{2} to both sides.

$$32x^{2}-18x+4x^{2}=3$$

Combine 32x^{2} and 4x^{2} to get 36x^{2}.

$$36x^{2}-18x=3$$

Divide both sides by 36.

$$\frac{36x^{2}-18x}{36}=\frac{3}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$x^{2}+\left(-\frac{18}{36}\right)x=\frac{3}{36}$$

Reduce the fraction \frac{-18}{36} to lowest terms by extracting and canceling out 18.

$$x^{2}-\frac{1}{2}x=\frac{3}{36}$$

Reduce the fraction \frac{3}{36} to lowest terms by extracting and canceling out 3.

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

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{1}{12}+\left(-\frac{1}{4}\right)^{2}$$

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{12}+\frac{1}{16}$$

Add \frac{1}{12} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{7}{48}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{7}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{7}{48}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{21}}{12} x-\frac{1}{4}=-\frac{\sqrt{21}}{12}$$

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

$$x=\frac{\sqrt{21}}{12}+\frac{1}{4} x=-\frac{\sqrt{21}}{12}+\frac{1}{4}$$