Solve -(x+1)(x+4)-x+3*x=8

$$- ( x + 1 ) ( x + 4 ) - x + 3 \cdot x = 8$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{39}i-3}{2}\approx -1.5-3.122498999i$$

$$x=\frac{-3+\sqrt{39}i}{2}\approx -1.5+3.122498999i$$

Steps Using the Quadratic Formula

To find the opposite of x+1, find the opposite of each term.

$$\left(-x-1\right)\left(x+4\right)-x+3x=8$$

Apply the distributive property by multiplying each term of -x-1 by each term of x+4.

$$-x^{2}-4x-x-4-x+3x=8$$

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

$$-x^{2}-5x-4-x+3x=8$$

Combine -5x and -x to get -6x.

$$-x^{2}-6x-4+3x=8$$

Combine -6x and 3x to get -3x.

$$-x^{2}-3x-4=8$$

Subtract 8 from both sides.

$$-x^{2}-3x-4-8=0$$

Subtract 8 from -4 to get -12.

$$-x^{2}-3x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$x=\frac{-\left(-3\right)±\sqrt{9-48}}{2\left(-1\right)}$$

Add 9 to -48.

$$x=\frac{-\left(-3\right)±\sqrt{-39}}{2\left(-1\right)}$$

Take the square root of -39.

$$x=\frac{-\left(-3\right)±\sqrt{39}i}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{39}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±\sqrt{39}i}{-2}$$

Now solve the equation x=\frac{3±\sqrt{39}i}{-2} when ± is plus. Add 3 to i\sqrt{39}.

$$x=\frac{3+\sqrt{39}i}{-2}$$

Divide 3+i\sqrt{39} by -2.

$$x=\frac{-\sqrt{39}i-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{39}i}{-2} when ± is minus. Subtract i\sqrt{39} from 3.

$$x=\frac{-\sqrt{39}i+3}{-2}$$

Divide 3-i\sqrt{39} by -2.

$$x=\frac{-3+\sqrt{39}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{39}i-3}{2} x=\frac{-3+\sqrt{39}i}{2}$$

Steps for Completing the Square

To find the opposite of x+1, find the opposite of each term.

$$\left(-x-1\right)\left(x+4\right)-x+3x=8$$

Apply the distributive property by multiplying each term of -x-1 by each term of x+4.

$$-x^{2}-4x-x-4-x+3x=8$$

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

$$-x^{2}-5x-4-x+3x=8$$

Combine -5x and -x to get -6x.

$$-x^{2}-6x-4+3x=8$$

Combine -6x and 3x to get -3x.

$$-x^{2}-3x-4=8$$

Add 4 to both sides.

$$-x^{2}-3x=8+4$$

Add 8 and 4 to get 12.

$$-x^{2}-3x=12$$

Divide both sides by -1.

$$\frac{-x^{2}-3x}{-1}=\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=\frac{12}{-1}$$

Divide -3 by -1.

$$x^{2}+3x=\frac{12}{-1}$$

Divide 12 by -1.

$$x^{2}+3x=-12$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-12+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=-12+\frac{9}{4}$$

Add -12 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{39}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{39}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{39}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{39}i}{2} x+\frac{3}{2}=-\frac{\sqrt{39}i}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{-3+\sqrt{39}i}{2} x=\frac{-\sqrt{39}i-3}{2}$$