Solve left(x+5right)^2=(1x)(1-x)

$${ \left(x+5 \right) }^{ 2 } =(1x)(1-x)$$

Solve for x (complex solution)

$$x=\frac{-9+\sqrt{119}i}{4}\approx -2.25+2.727178029i$$

$$x=\frac{-\sqrt{119}i-9}{4}\approx -2.25-2.727178029i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=1x\left(1-x\right)$$

Use the distributive property to multiply 1x by 1-x.

$$x^{2}+10x+25=1x-x^{2}$$

Subtract 1x from both sides.

$$x^{2}+10x+25-x=-x^{2}$$

Combine 10x and -x to get 9x.

$$x^{2}+9x+25=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}+9x+25+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+9x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 9 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 2\times 25}}{2\times 2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 2\times 25}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-9±\sqrt{81-8\times 25}}{2\times 2}$$

Multiply -8 times 25.

$$x=\frac{-9±\sqrt{81-200}}{2\times 2}$$

Add 81 to -200.

$$x=\frac{-9±\sqrt{-119}}{2\times 2}$$

Take the square root of -119.

$$x=\frac{-9±\sqrt{119}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-9±\sqrt{119}i}{4}$$

Now solve the equation x=\frac{-9±\sqrt{119}i}{4} when ± is plus. Add -9 to i\sqrt{119}.

$$x=\frac{-9+\sqrt{119}i}{4}$$

Now solve the equation x=\frac{-9±\sqrt{119}i}{4} when ± is minus. Subtract i\sqrt{119} from -9.

$$x=\frac{-\sqrt{119}i-9}{4}$$

The equation is now solved.

$$x=\frac{-9+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-9}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=1x\left(1-x\right)$$

Use the distributive property to multiply 1x by 1-x.

$$x^{2}+10x+25=1x-x^{2}$$

Subtract 1x from both sides.

$$x^{2}+10x+25-x=-x^{2}$$

Combine 10x and -x to get 9x.

$$x^{2}+9x+25=-x^{2}$$

Add x^{2} to both sides.

$$x^{2}+9x+25+x^{2}=0$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+9x+25=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+9x=-25$$

Divide both sides by 2.

$$\frac{2x^{2}+9x}{2}=-\frac{25}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{9}{2}x=-\frac{25}{2}$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{2}x+\left(\frac{9}{4}\right)^{2}=-\frac{25}{2}+\left(\frac{9}{4}\right)^{2}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=-\frac{25}{2}+\frac{81}{16}$$

Add -\frac{25}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=-\frac{119}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{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}.

$$\left(x+\frac{9}{4}\right)^{2}=-\frac{119}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{-\frac{119}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{\sqrt{119}i}{4} x+\frac{9}{4}=-\frac{\sqrt{119}i}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$x=\frac{-9+\sqrt{119}i}{4} x=\frac{-\sqrt{119}i-9}{4}$$