Solve 2x^2-3x^2+4x+4=0

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

Solve for x

$$x=2\sqrt{2}+2\approx 4.828427125$$

$$x=2-2\sqrt{2}\approx -0.828427125$$

Steps Using the Quadratic Formula

Combine 2x^{2} and -3x^{2} to get -x^{2}.

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

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

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-4±\sqrt{16+16}}{2\left(-1\right)}$$

Add 16 to 16.

$$x=\frac{-4±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of 32.

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

Multiply 2 times -1.

$$x=\frac{-4±4\sqrt{2}}{-2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Combine 2x^{2} and -3x^{2} to get -x^{2}.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 4 by -1.

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

Divide -4 by -1.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Add 4 to 4.

$$x^{2}-4x+4=8$$

Factor x^{2}-4x+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}.

$$\left(x-2\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{8}$$

Simplify.

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

Add 2 to both sides of the equation.

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