Solve (x-3)^2-(2x+1)^2=5

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

Solve for x

$$x=\frac{\sqrt{34}-5}{3}\approx 0.276983965$$

$$x=\frac{-\sqrt{34}-5}{3}\approx -3.610317298$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-\left(2x+1\right)^{2}=5$$

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

$$x^{2}-6x+9-\left(4x^{2}+4x+1\right)=5$$

To find the opposite of 4x^{2}+4x+1, find the opposite of each term.

$$x^{2}-6x+9-4x^{2}-4x-1=5$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-6x+9-4x-1=5$$

Combine -6x and -4x to get -10x.

$$-3x^{2}-10x+9-1=5$$

Subtract 1 from 9 to get 8.

$$-3x^{2}-10x+8=5$$

Subtract 5 from both sides.

$$-3x^{2}-10x+8-5=0$$

Subtract 5 from 8 to get 3.

$$-3x^{2}-10x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+12\times 3}}{2\left(-3\right)}$$

Multiply 12 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100+36}}{2\left(-3\right)}$$

Add 100 to 36.

$$x=\frac{-\left(-10\right)±\sqrt{136}}{2\left(-3\right)}$$

Take the square root of 136.

$$x=\frac{-\left(-10\right)±2\sqrt{34}}{2\left(-3\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{34}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{10±2\sqrt{34}}{-6}$$

Now solve the equation x=\frac{10±2\sqrt{34}}{-6} when ± is plus. Add 10 to 2\sqrt{34}.

$$x=\frac{2\sqrt{34}+10}{-6}$$

Divide 10+2\sqrt{34} by -6.

$$x=\frac{-\sqrt{34}-5}{3}$$

Now solve the equation x=\frac{10±2\sqrt{34}}{-6} when ± is minus. Subtract 2\sqrt{34} from 10.

$$x=\frac{10-2\sqrt{34}}{-6}$$

Divide 10-2\sqrt{34} by -6.

$$x=\frac{\sqrt{34}-5}{3}$$

The equation is now solved.

$$x=\frac{-\sqrt{34}-5}{3} x=\frac{\sqrt{34}-5}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9-\left(2x+1\right)^{2}=5$$

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

$$x^{2}-6x+9-\left(4x^{2}+4x+1\right)=5$$

To find the opposite of 4x^{2}+4x+1, find the opposite of each term.

$$x^{2}-6x+9-4x^{2}-4x-1=5$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-6x+9-4x-1=5$$

Combine -6x and -4x to get -10x.

$$-3x^{2}-10x+9-1=5$$

Subtract 1 from 9 to get 8.

$$-3x^{2}-10x+8=5$$

Subtract 8 from both sides.

$$-3x^{2}-10x=5-8$$

Subtract 8 from 5 to get -3.

$$-3x^{2}-10x=-3$$

Divide both sides by -3.

$$\frac{-3x^{2}-10x}{-3}=-\frac{3}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{10}{-3}\right)x=-\frac{3}{-3}$$

Divide -10 by -3.

$$x^{2}+\frac{10}{3}x=-\frac{3}{-3}$$

Divide -3 by -3.

$$x^{2}+\frac{10}{3}x=1$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=1+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=1+\frac{25}{9}$$

Add 1 to \frac{25}{9}.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{34}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{34}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{34}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{\sqrt{34}}{3} x+\frac{5}{3}=-\frac{\sqrt{34}}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=\frac{\sqrt{34}-5}{3} x=\frac{-\sqrt{34}-5}{3}$$