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

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

Solve for x

$$x=\frac{\sqrt{57}-5}{8}\approx 0.318729304$$

$$x=\frac{-\sqrt{57}-5}{8}\approx -1.568729304$$

Steps Using the Quadratic Formula

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

$$4x^{2}+4x+1=3-x$$

Subtract 3 from both sides.

$$4x^{2}+4x+1-3=-x$$

Subtract 3 from 1 to get -2.

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

Add x to both sides.

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

Combine 4x and x to get 5x.

$$4x^{2}+5x-2=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-2\right)}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-2\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\left(-2\right)}}{2\times 4}$$

Multiply -16 times -2.

$$x=\frac{-5±\sqrt{25+32}}{2\times 4}$$

Add 25 to 32.

$$x=\frac{-5±\sqrt{57}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±\sqrt{57}}{8}$$

Now solve the equation x=\frac{-5±\sqrt{57}}{8} when ± is plus. Add -5 to \sqrt{57}.

$$x=\frac{\sqrt{57}-5}{8}$$

Now solve the equation x=\frac{-5±\sqrt{57}}{8} when ± is minus. Subtract \sqrt{57} from -5.

$$x=\frac{-\sqrt{57}-5}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$

Steps for Completing the Square

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

$$4x^{2}+4x+1=3-x$$

Add x to both sides.

$$4x^{2}+4x+1+x=3$$

Combine 4x and x to get 5x.

$$4x^{2}+5x+1=3$$

Subtract 1 from both sides.

$$4x^{2}+5x=3-1$$

Subtract 1 from 3 to get 2.

$$4x^{2}+5x=2$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{4}x=\frac{1}{2}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{1}{2}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{1}{2}+\frac{25}{64}$$

Add \frac{1}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{57}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{57}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{57}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{57}}{8} x+\frac{5}{8}=-\frac{\sqrt{57}}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{\sqrt{57}-5}{8} x=\frac{-\sqrt{57}-5}{8}$$