Solve 3x-x^2/4=3

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{39}i+3}{2}\approx 1.5-3.122498999i$$

$$x=\frac{3+\sqrt{39}i}{2}\approx 1.5+3.122498999i$$

Steps Using the Quadratic Formula

Multiply both sides by 4.

$$3x-x^{2}=3\times 4$$

Multiply 3 and 4 to get 12.

$$3x-x^{2}=12$$

Subtract 12 from both sides.

$$3x-x^{2}-12=0$$

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.

$$-x^{2}+3x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-3±\sqrt{9+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-3±\sqrt{9-48}}{2\left(-1\right)}$$

Add 9 to -48.

$$x=\frac{-3±\sqrt{-39}}{2\left(-1\right)}$$

Take the square root of -39.

$$x=\frac{-3±\sqrt{39}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±\sqrt{39}i}{-2}$$

Now solve the equation x=\frac{-3±\sqrt{39}i}{-2} when ± is plus. Add -3 to i\sqrt{39}.

$$x=\frac{-3+\sqrt{39}i}{-2}$$

Divide -3+i\sqrt{39} by -2.

$$x=\frac{-\sqrt{39}i+3}{2}$$

Now solve the equation x=\frac{-3±\sqrt{39}i}{-2} when ± is minus. Subtract i\sqrt{39} from -3.

$$x=\frac{-\sqrt{39}i-3}{-2}$$

Divide -3-i\sqrt{39} by -2.

$$x=\frac{3+\sqrt{39}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{39}i+3}{2} x=\frac{3+\sqrt{39}i}{2}$$

Steps for Completing the Square

Multiply both sides by 4.

$$3x-x^{2}=3\times 4$$

Multiply 3 and 4 to get 12.

$$3x-x^{2}=12$$

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.

$$-x^{2}+3x=12$$

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

$$x^{2}-3x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}-3x=-12$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -12 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=-\frac{39}{4}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{39}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{39}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{39}i}{2} x-\frac{3}{2}=-\frac{\sqrt{39}i}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{3+\sqrt{39}i}{2} x=\frac{-\sqrt{39}i+3}{2}$$