Solve (8+x)*x=-6*6

$$( 8 + x ) \cdot x = - 6 \cdot 6$$

Solve for x (complex solution)

$$x=-4+2\sqrt{5}i\approx -4+4.472135955i$$

$$x=-2\sqrt{5}i-4\approx -4-4.472135955i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 8+x by x.

$$8x+x^{2}=-6\times 6$$

Multiply -6 and 6 to get -36.

$$8x+x^{2}=-36$$

Add 36 to both sides.

$$8x+x^{2}+36=0$$

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.

$$x^{2}+8x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 8 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 36}}{2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 36}}{2}$$

Multiply -4 times 36.

$$x=\frac{-8±\sqrt{64-144}}{2}$$

Add 64 to -144.

$$x=\frac{-8±\sqrt{-80}}{2}$$

Take the square root of -80.

$$x=\frac{-8±4\sqrt{5}i}{2}$$

Now solve the equation x=\frac{-8±4\sqrt{5}i}{2} when ± is plus. Add -8 to 4i\sqrt{5}.

$$x=\frac{-8+4\sqrt{5}i}{2}$$

Divide -8+4i\sqrt{5} by 2.

$$x=-4+2\sqrt{5}i$$

Now solve the equation x=\frac{-8±4\sqrt{5}i}{2} when ± is minus. Subtract 4i\sqrt{5} from -8.

$$x=\frac{-4\sqrt{5}i-8}{2}$$

Divide -8-4i\sqrt{5} by 2.

$$x=-2\sqrt{5}i-4$$

The equation is now solved.

$$x=-4+2\sqrt{5}i x=-2\sqrt{5}i-4$$

Steps for Completing the Square

Use the distributive property to multiply 8+x by x.

$$8x+x^{2}=-6\times 6$$

Multiply -6 and 6 to get -36.

$$8x+x^{2}=-36$$

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.

$$x^{2}+8x=-36$$

Divide 8, the coefficient of the x term, by 2 to get 4. Then add the square of 4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+8x+4^{2}=-36+4^{2}$$

Square 4.

$$x^{2}+8x+16=-36+16$$

Add -36 to 16.

$$x^{2}+8x+16=-20$$

Factor x^{2}+8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+4\right)^{2}=-20$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+4\right)^{2}}=\sqrt{-20}$$

Simplify.

$$x+4=2\sqrt{5}i x+4=-2\sqrt{5}i$$

Subtract 4 from both sides of the equation.

$$x=-4+2\sqrt{5}i x=-2\sqrt{5}i-4$$