Solve (x)=2x^2-3x+4

$$( x ) = 2 x ^ { 2 } - 3 x + 4$$

Solve for x (complex solution)

$$x=1+i$$

$$x=1-i$$

Steps Using the Quadratic Formula

Subtract 2x^{2} from both sides.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

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

Subtract 4 from both sides.

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

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.

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

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

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

Square 4.

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

Multiply -4 times -2.

$$x=\frac{-4±\sqrt{16+8\left(-4\right)}}{2\left(-2\right)}$$

Multiply 8 times -4.

$$x=\frac{-4±\sqrt{16-32}}{2\left(-2\right)}$$

Add 16 to -32.

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

Take the square root of -16.

$$x=\frac{-4±4i}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-4±4i}{-4}$$

Now solve the equation x=\frac{-4±4i}{-4} when ± is plus. Add -4 to 4i.

$$x=\frac{-4+4i}{-4}$$

Divide -4+4i by -4.

$$x=1-i$$

Now solve the equation x=\frac{-4±4i}{-4} when ± is minus. Subtract 4i from -4.

$$x=\frac{-4-4i}{-4}$$

Divide -4-4i by -4.

$$x=1+i$$

The equation is now solved.

$$x=1-i x=1+i$$

Steps for Completing the Square

Subtract 2x^{2} from both sides.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

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

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.

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

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 4 by -2.

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

Divide 4 by -2.

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

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

$$x^{2}-2x+1=-2+1$$

Add -2 to 1.

$$x^{2}-2x+1=-1$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=i x-1=-i$$

Add 1 to both sides of the equation.

$$x=1+i x=1-i$$