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

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

Solve for x (complex solution)

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

Steps Using the Quadratic Formula

Add 2x^{2} to both sides.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

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

Subtract 2 from both sides.

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

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.

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

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

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

Square 4.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

$$x=\frac{-4±\sqrt{16+16}}{2\times 2}$$

Add 16 to 16.

$$x=\frac{-4±\sqrt{32}}{2\times 2}$$

Take the square root of 32.

$$x=\frac{-4±2^{\frac{5}{2}}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±2^{\frac{5}{2}}}{4}$$

Now solve the equation x=\frac{-4±2^{\frac{5}{2}}}{4} when ± is plus. Add -4 to 2^{\frac{5}{2}}.

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

Divide -4+2^{\frac{5}{2}} by 4.

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

Now solve the equation x=\frac{-4±2^{\frac{5}{2}}}{4} when ± is minus. Subtract 2^{\frac{5}{2}} from -4.

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

Divide -4-4\sqrt{2} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

Add 2x^{2} to both sides.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

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

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.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

Divide 2 by 2.

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

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.

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

Square 1.

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

Add 1 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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

Solve for x

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\sqrt{2}-1\approx -2.414213562$$

Steps Using the Quadratic Formula

Add 2x^{2} to both sides.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

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

Subtract 2 from both sides.

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

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.

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

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

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

Square 4.

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

Multiply -4 times 2.

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

Multiply -8 times -2.

$$x=\frac{-4±\sqrt{16+16}}{2\times 2}$$

Add 16 to 16.

$$x=\frac{-4±\sqrt{32}}{2\times 2}$$

Take the square root of 32.

$$x=\frac{-4±4\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-4±4\sqrt{2}}{4}$$

Now solve the equation x=\frac{-4±4\sqrt{2}}{4} when ± is plus. Add -4 to 4\sqrt{2}.

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

Divide -4+4\sqrt{2} by 4.

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

Now solve the equation x=\frac{-4±4\sqrt{2}}{4} when ± is minus. Subtract 4\sqrt{2} from -4.

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

Divide -4-4\sqrt{2} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

Add 2x^{2} to both sides.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

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

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.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

Divide 2 by 2.

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

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.

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

Square 1.

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

Add 1 to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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