Solve x=left(x-9div2right)^2-9div4

$$x= { \left(x-9 \div 2 \right) }^{ 2 } -9 \div 4$$

Solve for x

$$x=\sqrt{7}+5\approx 7.645751311$$

$$x=5-\sqrt{7}\approx 2.354248689$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{9}{2}\right)^{2}.

$$x=x^{2}-9x+\frac{81}{4}-\frac{9}{4}$$

Subtract \frac{9}{4} from \frac{81}{4} to get 18.

$$x=x^{2}-9x+18$$

Subtract x^{2} from both sides.

$$x-x^{2}=-9x+18$$

Add 9x to both sides.

$$x-x^{2}+9x=18$$

Combine x and 9x to get 10x.

$$10x-x^{2}=18$$

Subtract 18 from both sides.

$$10x-x^{2}-18=0$$

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.

$$-x^{2}+10x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-18\right)}}{2\left(-1\right)}$$

Multiply 4 times -18.

$$x=\frac{-10±\sqrt{100-72}}{2\left(-1\right)}$$

Add 100 to -72.

$$x=\frac{-10±\sqrt{28}}{2\left(-1\right)}$$

Take the square root of 28.

$$x=\frac{-10±2\sqrt{7}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±2\sqrt{7}}{-2}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{-2} when ± is plus. Add -10 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-10}{-2}$$

Divide -10+2\sqrt{7} by -2.

$$x=5-\sqrt{7}$$

Now solve the equation x=\frac{-10±2\sqrt{7}}{-2} when ± is minus. Subtract 2\sqrt{7} from -10.

$$x=\frac{-2\sqrt{7}-10}{-2}$$

Divide -10-2\sqrt{7} by -2.

$$x=\sqrt{7}+5$$

The equation is now solved.

$$x=5-\sqrt{7} x=\sqrt{7}+5$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-\frac{9}{2}\right)^{2}.

$$x=x^{2}-9x+\frac{81}{4}-\frac{9}{4}$$

Subtract \frac{9}{4} from \frac{81}{4} to get 18.

$$x=x^{2}-9x+18$$

Subtract x^{2} from both sides.

$$x-x^{2}=-9x+18$$

Add 9x to both sides.

$$x-x^{2}+9x=18$$

Combine x and 9x to get 10x.

$$10x-x^{2}=18$$

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.

$$-x^{2}+10x=18$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{18}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{18}{-1}$$

Divide 18 by -1.

$$x^{2}-10x=-18$$

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

$$x^{2}-10x+\left(-5\right)^{2}=-18+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-18+25$$

Add -18 to 25.

$$x^{2}-10x+25=7$$

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

$$\left(x-5\right)^{2}=7$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{7}$$

Simplify.

$$x-5=\sqrt{7} x-5=-\sqrt{7}$$

Add 5 to both sides of the equation.

$$x=\sqrt{7}+5 x=5-\sqrt{7}$$