Solve 1/2x^2=1/3(x-1)+(x-3)(x+4)/12

$$\frac { 1 } { 2 } x ^ { 2 } = \frac { 1 } { 3 } ( x - 1 ) + \frac { ( x - 3 ) ( x + 4 ) } { 12 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{295}i}{10}+\frac{1}{2}\approx 0.5+1.717556404i$$

$$x=-\frac{\sqrt{295}i}{10}+\frac{1}{2}\approx 0.5-1.717556404i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 2,3,12.

$$6x^{2}=4\left(x-1\right)+\left(x-3\right)\left(x+4\right)$$

Use the distributive property to multiply 4 by x-1.

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

Use the distributive property to multiply x-3 by x+4 and combine like terms.

$$6x^{2}=4x-4+x^{2}+x-12$$

Combine 4x and x to get 5x.

$$6x^{2}=5x-4+x^{2}-12$$

Subtract 12 from -4 to get -16.

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

Subtract 5x from both sides.

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

Subtract -16 from both sides.

$$6x^{2}-5x-\left(-16\right)=x^{2}$$

The opposite of -16 is 16.

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

Subtract x^{2} from both sides.

$$6x^{2}-5x+16-x^{2}=0$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-5x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -5 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 5\times 16}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-5\right)±\sqrt{25-20\times 16}}{2\times 5}$$

Multiply -20 times 16.

$$x=\frac{-\left(-5\right)±\sqrt{25-320}}{2\times 5}$$

Add 25 to -320.

$$x=\frac{-\left(-5\right)±\sqrt{-295}}{2\times 5}$$

Take the square root of -295.

$$x=\frac{-\left(-5\right)±\sqrt{295}i}{2\times 5}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{295}i}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{5±\sqrt{295}i}{10}$$

Now solve the equation x=\frac{5±\sqrt{295}i}{10} when ± is plus. Add 5 to i\sqrt{295}.

$$x=\frac{5+\sqrt{295}i}{10}$$

Divide 5+i\sqrt{295} by 10.

$$x=\frac{\sqrt{295}i}{10}+\frac{1}{2}$$

Now solve the equation x=\frac{5±\sqrt{295}i}{10} when ± is minus. Subtract i\sqrt{295} from 5.

$$x=\frac{-\sqrt{295}i+5}{10}$$

Divide 5-i\sqrt{295} by 10.

$$x=-\frac{\sqrt{295}i}{10}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{295}i}{10}+\frac{1}{2} x=-\frac{\sqrt{295}i}{10}+\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 2,3,12.

$$6x^{2}=4\left(x-1\right)+\left(x-3\right)\left(x+4\right)$$

Use the distributive property to multiply 4 by x-1.

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

Use the distributive property to multiply x-3 by x+4 and combine like terms.

$$6x^{2}=4x-4+x^{2}+x-12$$

Combine 4x and x to get 5x.

$$6x^{2}=5x-4+x^{2}-12$$

Subtract 12 from -4 to get -16.

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

Subtract 5x from both sides.

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

Subtract x^{2} from both sides.

$$6x^{2}-5x-x^{2}=-16$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$5x^{2}-5x=-16$$

Divide both sides by 5.

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

Dividing by 5 undoes the multiplication by 5.

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

Divide -5 by 5.

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

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{16}{5}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{16}{5}+\frac{1}{4}$$

Add -\frac{16}{5} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{59}{20}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{59}{20}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{59}{20}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{295}i}{10} x-\frac{1}{2}=-\frac{\sqrt{295}i}{10}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{295}i}{10}+\frac{1}{2} x=-\frac{\sqrt{295}i}{10}+\frac{1}{2}$$