Solve 33-x^2-2left(x+1right)^2=0

$$33- { x }^{ 2 } -2 { \left(x+1 \right) }^{ 2 } =0$$

Solve for x

$$x = \frac{\sqrt{97} - 2}{3} \approx 2.616285934$$

$$x=\frac{-\sqrt{97}-2}{3}\approx -3.949619267$$

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}.

$$33-x^{2}-2\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -2 by x^{2}+2x+1.

$$33-x^{2}-2x^{2}-4x-2=0$$

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

$$33-3x^{2}-4x-2=0$$

Subtract 2 from 33 to get 31.

$$31-3x^{2}-4x=0$$

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.

$$-3x^{2}-4x+31=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -4 for b, and 31 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-3\right)\times 31}}{2\left(-3\right)}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-3\right)\times 31}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-4\right)±\sqrt{16+12\times 31}}{2\left(-3\right)}$$

Multiply 12 times 31.

$$x=\frac{-\left(-4\right)±\sqrt{16+372}}{2\left(-3\right)}$$

Add 16 to 372.

$$x=\frac{-\left(-4\right)±\sqrt{388}}{2\left(-3\right)}$$

Take the square root of 388.

$$x=\frac{-\left(-4\right)±2\sqrt{97}}{2\left(-3\right)}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{97}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{4±2\sqrt{97}}{-6}$$

Now solve the equation x=\frac{4±2\sqrt{97}}{-6} when ± is plus. Add 4 to 2\sqrt{97}.

$$x=\frac{2\sqrt{97}+4}{-6}$$

Divide 4+2\sqrt{97} by -6.

$$x=\frac{-\sqrt{97}-2}{3}$$

Now solve the equation x=\frac{4±2\sqrt{97}}{-6} when ± is minus. Subtract 2\sqrt{97} from 4.

$$x=\frac{4-2\sqrt{97}}{-6}$$

Divide 4-2\sqrt{97} by -6.

$$x=\frac{\sqrt{97}-2}{3}$$

The equation is now solved.

$$x=\frac{-\sqrt{97}-2}{3} x=\frac{\sqrt{97}-2}{3}$$

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}.

$$33-x^{2}-2\left(x^{2}+2x+1\right)=0$$

Use the distributive property to multiply -2 by x^{2}+2x+1.

$$33-x^{2}-2x^{2}-4x-2=0$$

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

$$33-3x^{2}-4x-2=0$$

Subtract 2 from 33 to get 31.

$$31-3x^{2}-4x=0$$

Subtract 31 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}-4x=-31$$

Divide both sides by -3.

$$\frac{-3x^{2}-4x}{-3}=-\frac{31}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{4}{-3}\right)x=-\frac{31}{-3}$$

Divide -4 by -3.

$$x^{2}+\frac{4}{3}x=-\frac{31}{-3}$$

Divide -31 by -3.

$$x^{2}+\frac{4}{3}x=\frac{31}{3}$$

Divide \frac{4}{3}, the coefficient of the x term, by 2 to get \frac{2}{3}. Then add the square of \frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{31}{3}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{31}{3}+\frac{4}{9}$$

Add \frac{31}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{97}{9}$$

Factor x^{2}+\frac{4}{3}x+\frac{4}{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}.

$$\left(x+\frac{2}{3}\right)^{2}=\frac{97}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{97}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{\sqrt{97}}{3} x+\frac{2}{3}=-\frac{\sqrt{97}}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=\frac{\sqrt{97}-2}{3} x=\frac{-\sqrt{97}-2}{3}$$