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

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

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{2}i}{2}\approx -0.5+0.707106781i$$

$$x=\frac{-\sqrt{2}i-1}{2}\approx -0.5-0.707106781i$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

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

Subtract 3x from both sides.

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

Combine 2x and -3x to get -x.

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

Add \frac{9}{4} to both sides.

$$-x-3-x^{2}+\frac{9}{4}=0$$

Add -3 and \frac{9}{4} to get -\frac{3}{4}.

$$-x-\frac{3}{4}-x^{2}=0$$

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.

$$-x^{2}-x-\frac{3}{4}=0$$

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

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

Multiply -4 times -1.

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

Multiply 4 times -\frac{3}{4}.

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

Add 1 to -3.

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

Take the square root of -2.

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

The opposite of -1 is 1.

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

Multiply 2 times -1.

$$x=\frac{1±\sqrt{2}i}{-2}$$

Now solve the equation x=\frac{1±\sqrt{2}i}{-2} when ± is plus. Add 1 to i\sqrt{2}.

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

Divide 1+i\sqrt{2} by -2.

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

Now solve the equation x=\frac{1±\sqrt{2}i}{-2} when ± is minus. Subtract i\sqrt{2} from 1.

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

Divide 1-i\sqrt{2} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 3x from both sides.

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

Combine 2x and -3x to get -x.

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

Add 3 to both sides.

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

Add -\frac{9}{4} and 3 to get \frac{3}{4}.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=\frac{\frac{3}{4}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide \frac{3}{4} by -1.

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

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.

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

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

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

Add -\frac{3}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

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