Solve 10x+27=-2-x-x^2

$$10 x + 27 = - 2 - x - x ^ { 2 }$$

Solve for x

$$x=\frac{\sqrt{5}-11}{2}\approx -4.381966011$$

$$x=\frac{-\sqrt{5}-11}{2}\approx -6.618033989$$

Steps Using the Quadratic Formula

Subtract -2 from both sides.

$$10x+27-\left(-2\right)=-x-x^{2}$$

The opposite of -2 is 2.

$$10x+27+2=-x-x^{2}$$

Add x to both sides.

$$10x+27+2+x=-x^{2}$$

Add 27 and 2 to get 29.

$$10x+29+x=-x^{2}$$

Combine 10x and x to get 11x.

$$11x+29=-x^{2}$$

Add x^{2} to both sides.

$$11x+29+x^{2}=0$$

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.

$$x^{2}+11x+29=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 11 for b, and 29 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 29}}{2}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 29}}{2}$$

Multiply -4 times 29.

$$x=\frac{-11±\sqrt{121-116}}{2}$$

Add 121 to -116.

$$x=\frac{-11±\sqrt{5}}{2}$$

Now solve the equation x=\frac{-11±\sqrt{5}}{2} when ± is plus. Add -11 to \sqrt{5}.

$$x=\frac{\sqrt{5}-11}{2}$$

Now solve the equation x=\frac{-11±\sqrt{5}}{2} when ± is minus. Subtract \sqrt{5} from -11.

$$x=\frac{-\sqrt{5}-11}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{5}-11}{2} x=\frac{-\sqrt{5}-11}{2}$$

Steps for Completing the Square

Add x to both sides.

$$10x+27+x=-2-x^{2}$$

Combine 10x and x to get 11x.

$$11x+27=-2-x^{2}$$

Add x^{2} to both sides.

$$11x+27+x^{2}=-2$$

Subtract 27 from both sides.

$$11x+x^{2}=-2-27$$

Subtract 27 from -2 to get -29.

$$11x+x^{2}=-29$$

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.

$$x^{2}+11x=-29$$

Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=-29+\left(\frac{11}{2}\right)^{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+11x+\frac{121}{4}=-29+\frac{121}{4}$$

Add -29 to \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{5}{4}$$

Factor x^{2}+11x+\frac{121}{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}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{5}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Simplify.

$$x+\frac{11}{2}=\frac{\sqrt{5}}{2} x+\frac{11}{2}=-\frac{\sqrt{5}}{2}$$

Subtract \frac{11}{2} from both sides of the equation.

$$x=\frac{\sqrt{5}-11}{2} x=\frac{-\sqrt{5}-11}{2}$$