Solve left(x+1right)^2=1-3x

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

Solve for x

$$x=-5$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Add 3x to both sides.

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

Combine 2x and 3x to get 5x.

$$x^{2}+5x=0$$

Factor out x.

$$x\left(x+5\right)=0$$

To find equation solutions, solve x=0 and x+5=0.

$$x=0 x=-5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Add 3x to both sides.

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

Combine 2x and 3x to get 5x.

$$x^{2}+5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2}$$

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

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

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

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=0 x=-5$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

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

Add 3x to both sides.

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

Combine 2x and 3x to get 5x.

$$x^{2}+5x=0$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=\frac{25}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

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

Subtract \frac{5}{2} from both sides of the equation.

$$x=0 x=-5$$