Solve x+7=-x^2-3x+x+3

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

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{7}i}{2}\approx -1.5+1.322875656i$$

$$x=\frac{-\sqrt{7}i-3}{2}\approx -1.5-1.322875656i$$

Steps Using the Quadratic Formula

Combine -3x and x to get -2x.

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

Add x^{2} to both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

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

Subtract 3 from both sides.

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

Subtract 3 from 7 to get 4.

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

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.

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

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 4}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-3±\sqrt{9-16}}{2}$$

Add 9 to -16.

$$x=\frac{-3±\sqrt{-7}}{2}$$

Take the square root of -7.

$$x=\frac{-3±\sqrt{7}i}{2}$$

Now solve the equation x=\frac{-3±\sqrt{7}i}{2} when ± is plus. Add -3 to i\sqrt{7}.

$$x=\frac{-3+\sqrt{7}i}{2}$$

Now solve the equation x=\frac{-3±\sqrt{7}i}{2} when ± is minus. Subtract i\sqrt{7} from -3.

$$x=\frac{-\sqrt{7}i-3}{2}$$

The equation is now solved.

$$x=\frac{-3+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-3}{2}$$

Steps for Completing the Square

Combine -3x and x to get -2x.

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

Add x^{2} to both sides.

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

Add 2x to both sides.

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

Combine x and 2x to get 3x.

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

Subtract 7 from both sides.

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

Subtract 7 from 3 to get -4.

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

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.

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

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.

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

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

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

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

$$x^{2}+3x+\frac{9}{4}=-\frac{7}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{7}i}{2} x+\frac{3}{2}=-\frac{\sqrt{7}i}{2}$$

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

$$x=\frac{-3+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-3}{2}$$