Solve (7-x)[x-3)=1

$$( 7 - x ) [ x - 3 ) = 1$$

Solve for x

$$x=\sqrt{3}+5\approx 6.732050808$$

$$x=5-\sqrt{3}\approx 3.267949192$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 7-x by x-3 and combine like terms.

$$10x-21-x^{2}=1$$

Subtract 1 from both sides.

$$10x-21-x^{2}-1=0$$

Subtract 1 from -21 to get -22.

$$10x-22-x^{2}=0$$

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.

$$-x^{2}+10x-22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-22\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-22\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-22\right)}}{2\left(-1\right)}$$

Multiply 4 times -22.

$$x=\frac{-10±\sqrt{100-88}}{2\left(-1\right)}$$

Add 100 to -88.

$$x=\frac{-10±\sqrt{12}}{2\left(-1\right)}$$

Take the square root of 12.

$$x=\frac{-10±2\sqrt{3}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±2\sqrt{3}}{-2}$$

Now solve the equation x=\frac{-10±2\sqrt{3}}{-2} when ± is plus. Add -10 to 2\sqrt{3}.

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

Divide -10+2\sqrt{3} by -2.

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

Now solve the equation x=\frac{-10±2\sqrt{3}}{-2} when ± is minus. Subtract 2\sqrt{3} from -10.

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

Divide -10-2\sqrt{3} by -2.

$$x=\sqrt{3}+5$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 7-x by x-3 and combine like terms.

$$10x-21-x^{2}=1$$

Add 21 to both sides.

$$10x-x^{2}=1+21$$

Add 1 and 21 to get 22.

$$10x-x^{2}=22$$

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.

$$-x^{2}+10x=22$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{22}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{22}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{22}{-1}$$

Divide 22 by -1.

$$x^{2}-10x=-22$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=-22+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-22+25$$

Add -22 to 25.

$$x^{2}-10x+25=3$$

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}.

$$\left(x-5\right)^{2}=3$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{3}$$

Simplify.

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

Add 5 to both sides of the equation.

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