Solve 8-x^2+10-left(x+1right)^2=0

$$8- { x }^{ 2 } +10- { \left(x+1 \right) }^{ 2 } =0$$

Solve for x

$$x = \frac{\sqrt{35} - 1}{2} \approx 2.458039892$$

$$x=\frac{-\sqrt{35}-1}{2}\approx -3.458039892$$

Steps Using the Quadratic Formula

Add 8 and 10 to get 18.

$$18-x^{2}-\left(x+1\right)^{2}=0$$

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

$$18-x^{2}-\left(x^{2}+2x+1\right)=0$$

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

$$18-x^{2}-x^{2}-2x-1=0$$

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

$$18-2x^{2}-2x-1=0$$

Subtract 1 from 18 to get 17.

$$17-2x^{2}-2x=0$$

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.

$$-2x^{2}-2x+17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -2 for b, and 17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)\times 17}}{2\left(-2\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)\times 17}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8\times 17}}{2\left(-2\right)}$$

Multiply 8 times 17.

$$x=\frac{-\left(-2\right)±\sqrt{4+136}}{2\left(-2\right)}$$

Add 4 to 136.

$$x=\frac{-\left(-2\right)±\sqrt{140}}{2\left(-2\right)}$$

Take the square root of 140.

$$x=\frac{-\left(-2\right)±2\sqrt{35}}{2\left(-2\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{35}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{2±2\sqrt{35}}{-4}$$

Now solve the equation x=\frac{2±2\sqrt{35}}{-4} when ± is plus. Add 2 to 2\sqrt{35}.

$$x=\frac{2\sqrt{35}+2}{-4}$$

Divide 2+2\sqrt{35} by -4.

$$x=\frac{-\sqrt{35}-1}{2}$$

Now solve the equation x=\frac{2±2\sqrt{35}}{-4} when ± is minus. Subtract 2\sqrt{35} from 2.

$$x=\frac{2-2\sqrt{35}}{-4}$$

Divide 2-2\sqrt{35} by -4.

$$x=\frac{\sqrt{35}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{35}-1}{2} x=\frac{\sqrt{35}-1}{2}$$

Steps for Completing the Square

Add 8 and 10 to get 18.

$$18-x^{2}-\left(x+1\right)^{2}=0$$

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

$$18-x^{2}-\left(x^{2}+2x+1\right)=0$$

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

$$18-x^{2}-x^{2}-2x-1=0$$

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

$$18-2x^{2}-2x-1=0$$

Subtract 1 from 18 to get 17.

$$17-2x^{2}-2x=0$$

Subtract 17 from both sides. Anything subtracted from zero gives its negation.

$$-2x^{2}-2x=-17$$

Divide both sides by -2.

$$\frac{-2x^{2}-2x}{-2}=-\frac{17}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{2}{-2}\right)x=-\frac{17}{-2}$$

Divide -2 by -2.

$$x^{2}+x=-\frac{17}{-2}$$

Divide -17 by -2.

$$x^{2}+x=\frac{17}{2}$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{17}{2}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{17}{2}+\frac{1}{4}$$

Add \frac{17}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{35}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{35}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{35}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{35}}{2} x+\frac{1}{2}=-\frac{\sqrt{35}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{35}-1}{2} x=\frac{-\sqrt{35}-1}{2}$$