Solve x^2+3+8x=2x-1

$$x ^ { 2 } + 3 + 8 x = 2 x - 1$$

Solve for x (complex solution)

$$x=\sqrt{5}-3\approx -0.763932023$$

$$x=-\left(\sqrt{5}+3\right)\approx -5.236067977$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$x^{2}+3+8x-2x=-1$$

Combine 8x and -2x to get 6x.

$$x^{2}+3+6x=-1$$

Add 1 to both sides.

$$x^{2}+3+6x+1=0$$

Add 3 and 1 to get 4.

$$x^{2}+4+6x=0$$

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.

$$x^{2}+6x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16}}{2}$$

Add 36 to -16.

$$x=\frac{-6±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-6±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{2} when ± is plus. Add -6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-6}{2}$$

Divide -6+2\sqrt{5} by 2.

$$x=\sqrt{5}-3$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -6.

$$x=\frac{-2\sqrt{5}-6}{2}$$

Divide -6-2\sqrt{5} by 2.

$$x=-\sqrt{5}-3$$

The equation is now solved.

$$x=\sqrt{5}-3 x=-\sqrt{5}-3$$

Steps for Completing the Square

Subtract 2x from both sides.

$$x^{2}+3+8x-2x=-1$$

Combine 8x and -2x to get 6x.

$$x^{2}+3+6x=-1$$

Subtract 3 from both sides.

$$x^{2}+6x=-1-3$$

Subtract 3 from -1 to get -4.

$$x^{2}+6x=-4$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-4+3^{2}$$

Square 3.

$$x^{2}+6x+9=-4+9$$

Add -4 to 9.

$$x^{2}+6x+9=5$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+3=\sqrt{5} x+3=-\sqrt{5}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{5}-3 x=-\sqrt{5}-3$$

Solve for x

$$x=\sqrt{5}-3\approx -0.763932023$$

$$x=-\sqrt{5}-3\approx -5.236067977$$

Steps Using the Quadratic Formula

Subtract 2x from both sides.

$$x^{2}+3+8x-2x=-1$$

Combine 8x and -2x to get 6x.

$$x^{2}+3+6x=-1$$

Add 1 to both sides.

$$x^{2}+3+6x+1=0$$

Add 3 and 1 to get 4.

$$x^{2}+4+6x=0$$

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.

$$x^{2}+6x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 4}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-6±\sqrt{36-16}}{2}$$

Add 36 to -16.

$$x=\frac{-6±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{-6±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{2} when ± is plus. Add -6 to 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-6}{2}$$

Divide -6+2\sqrt{5} by 2.

$$x=\sqrt{5}-3$$

Now solve the equation x=\frac{-6±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from -6.

$$x=\frac{-2\sqrt{5}-6}{2}$$

Divide -6-2\sqrt{5} by 2.

$$x=-\sqrt{5}-3$$

The equation is now solved.

$$x=\sqrt{5}-3 x=-\sqrt{5}-3$$

Steps for Completing the Square

Subtract 2x from both sides.

$$x^{2}+3+8x-2x=-1$$

Combine 8x and -2x to get 6x.

$$x^{2}+3+6x=-1$$

Subtract 3 from both sides.

$$x^{2}+6x=-1-3$$

Subtract 3 from -1 to get -4.

$$x^{2}+6x=-4$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-4+3^{2}$$

Square 3.

$$x^{2}+6x+9=-4+9$$

Add -4 to 9.

$$x^{2}+6x+9=5$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+3=\sqrt{5} x+3=-\sqrt{5}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{5}-3 x=-\sqrt{5}-3$$