Solve x=1+left(x^2-32+6right)/2+1

$$x = 1+ \frac{ \left( { x }^{ 2 } -32+6 \right) }{ 2 } +1$$

Solve for x

$$x=\sqrt{23}+1\approx 5.795831523$$

$$x=1-\sqrt{23}\approx -3.795831523$$

Steps Using the Quadratic Formula

Add -32 and 6 to get -26.

$$x=1+\frac{x^{2}-26}{2}+1$$

Add 1 and 1 to get 2.

$$x=2+\frac{x^{2}-26}{2}$$

Divide each term of x^{2}-26 by 2 to get \frac{1}{2}x^{2}-13.

$$x=2+\frac{1}{2}x^{2}-13$$

Subtract 13 from 2 to get -11.

$$x=-11+\frac{1}{2}x^{2}$$

Subtract -11 from both sides.

$$x-\left(-11\right)=\frac{1}{2}x^{2}$$

The opposite of -11 is 11.

$$x+11=\frac{1}{2}x^{2}$$

Subtract \frac{1}{2}x^{2} from both sides.

$$x+11-\frac{1}{2}x^{2}=0$$

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.

$$-\frac{1}{2}x^{2}+x+11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, 1 for b, and 11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-\frac{1}{2}\right)\times 11}}{2\left(-\frac{1}{2}\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-\frac{1}{2}\right)\times 11}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-1±\sqrt{1+2\times 11}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times 11.

$$x=\frac{-1±\sqrt{1+22}}{2\left(-\frac{1}{2}\right)}$$

Add 1 to 22.

$$x=\frac{-1±\sqrt{23}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{-1±\sqrt{23}}{-1}$$

Now solve the equation x=\frac{-1±\sqrt{23}}{-1} when ± is plus. Add -1 to \sqrt{23}.

$$x=\frac{\sqrt{23}-1}{-1}$$

Divide -1+\sqrt{23} by -1.

$$x=1-\sqrt{23}$$

Now solve the equation x=\frac{-1±\sqrt{23}}{-1} when ± is minus. Subtract \sqrt{23} from -1.

$$x=\frac{-\sqrt{23}-1}{-1}$$

Divide -1-\sqrt{23} by -1.

$$x=\sqrt{23}+1$$

The equation is now solved.

$$x=1-\sqrt{23} x=\sqrt{23}+1$$

Steps for Completing the Square

Add -32 and 6 to get -26.

$$x=1+\frac{x^{2}-26}{2}+1$$

Add 1 and 1 to get 2.

$$x=2+\frac{x^{2}-26}{2}$$

Divide each term of x^{2}-26 by 2 to get \frac{1}{2}x^{2}-13.

$$x=2+\frac{1}{2}x^{2}-13$$

Subtract 13 from 2 to get -11.

$$x=-11+\frac{1}{2}x^{2}$$

Subtract \frac{1}{2}x^{2} from both sides.

$$x-\frac{1}{2}x^{2}=-11$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-\frac{1}{2}x^{2}+x=-11$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}x^{2}+x}{-\frac{1}{2}}=-\frac{11}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$x^{2}+\frac{1}{-\frac{1}{2}}x=-\frac{11}{-\frac{1}{2}}$$

Divide 1 by -\frac{1}{2} by multiplying 1 by the reciprocal of -\frac{1}{2}.

$$x^{2}-2x=-\frac{11}{-\frac{1}{2}}$$

Divide -11 by -\frac{1}{2} by multiplying -11 by the reciprocal of -\frac{1}{2}.

$$x^{2}-2x=22$$

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

$$x^{2}-2x+1=22+1$$

Add 22 to 1.

$$x^{2}-2x+1=23$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=23$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{23}$$

Simplify.

$$x-1=\sqrt{23} x-1=-\sqrt{23}$$

Add 1 to both sides of the equation.

$$x=\sqrt{23}+1 x=1-\sqrt{23}$$