Solve 2x^2+x+7=-x^2+5x+(-2-x^2)

$$2 x ^ { 2 } + x + 7 = - x ^ { 2 } + 5 x + ( - 2 - x ^ { 2 } )$$

Solve for x (complex solution)

$$x=\frac{1}{2}+\sqrt{2}i\approx 0.5+1.414213562i$$

$$x=-\sqrt{2}i+\frac{1}{2}\approx 0.5-1.414213562i$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$2x^{2}+x+7+x^{2}=5x-2-x^{2}$$

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

$$3x^{2}+x+7=5x-2-x^{2}$$

Subtract 5x from both sides.

$$3x^{2}+x+7-5x=-2-x^{2}$$

Combine x and -5x to get -4x.

$$3x^{2}-4x+7=-2-x^{2}$$

Subtract -2 from both sides.

$$3x^{2}-4x+7-\left(-2\right)=-x^{2}$$

The opposite of -2 is 2.

$$3x^{2}-4x+7+2=-x^{2}$$

Add x^{2} to both sides.

$$3x^{2}-4x+7+2+x^{2}=0$$

Add 7 and 2 to get 9.

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

Combine 3x^{2} and x^{2} to get 4x^{2}.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\times 9}}{2\times 4}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 4\times 9}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-4\right)±\sqrt{16-16\times 9}}{2\times 4}$$

Multiply -16 times 9.

$$x=\frac{-\left(-4\right)±\sqrt{16-144}}{2\times 4}$$

Add 16 to -144.

$$x=\frac{-\left(-4\right)±\sqrt{-128}}{2\times 4}$$

Take the square root of -128.

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

The opposite of -4 is 4.

$$x=\frac{4±8\sqrt{2}i}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{4±8\sqrt{2}i}{8}$$

Now solve the equation x=\frac{4±8\sqrt{2}i}{8} when ± is plus. Add 4 to 8i\sqrt{2}.

$$x=\frac{4+8\sqrt{2}i}{8}$$

Divide 4+8i\sqrt{2} by 8.

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

Now solve the equation x=\frac{4±8\sqrt{2}i}{8} when ± is minus. Subtract 8i\sqrt{2} from 4.

$$x=\frac{-8\sqrt{2}i+4}{8}$$

Divide 4-8i\sqrt{2} by 8.

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

The equation is now solved.

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

Steps for Completing the Square

Add x^{2} to both sides.

$$2x^{2}+x+7+x^{2}=5x-2-x^{2}$$

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

$$3x^{2}+x+7=5x-2-x^{2}$$

Subtract 5x from both sides.

$$3x^{2}+x+7-5x=-2-x^{2}$$

Combine x and -5x to get -4x.

$$3x^{2}-4x+7=-2-x^{2}$$

Add x^{2} to both sides.

$$3x^{2}-4x+7+x^{2}=-2$$

Combine 3x^{2} and x^{2} to get 4x^{2}.

$$4x^{2}-4x+7=-2$$

Subtract 7 from both sides.

$$4x^{2}-4x=-2-7$$

Subtract 7 from -2 to get -9.

$$4x^{2}-4x=-9$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{4}{4}\right)x=-\frac{9}{4}$$

Divide -4 by 4.

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

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{9}{4}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{-9+1}{4}$$

Add -\frac{9}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{2} to both sides of the equation.

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