Solve (5-x)*(3-x)=10

$$(5-x) \times (3-x)=10$$

Solve for x

$$x=4-\sqrt{11}\approx 0.68337521$$

$$x=\sqrt{11}+4\approx 7.31662479$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5-x by 3-x and combine like terms.

$$15-8x+x^{2}=10$$

Subtract 10 from both sides.

$$15-8x+x^{2}-10=0$$

Subtract 10 from 15 to get 5.

$$5-8x+x^{2}=0$$

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.

$$x^{2}-8x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5}}{2}$$

Multiply -4 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-20}}{2}$$

Add 64 to -20.

$$x=\frac{-\left(-8\right)±\sqrt{44}}{2}$$

Take the square root of 44.

$$x=\frac{-\left(-8\right)±2\sqrt{11}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{11}}{2}$$

Now solve the equation x=\frac{8±2\sqrt{11}}{2} when ± is plus. Add 8 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}+8}{2}$$

Divide 8+2\sqrt{11} by 2.

$$x=\sqrt{11}+4$$

Now solve the equation x=\frac{8±2\sqrt{11}}{2} when ± is minus. Subtract 2\sqrt{11} from 8.

$$x=\frac{8-2\sqrt{11}}{2}$$

Divide 8-2\sqrt{11} by 2.

$$x=4-\sqrt{11}$$

The equation is now solved.

$$x=\sqrt{11}+4 x=4-\sqrt{11}$$

Steps for Completing the Square

Use the distributive property to multiply 5-x by 3-x and combine like terms.

$$15-8x+x^{2}=10$$

Subtract 15 from both sides.

$$-8x+x^{2}=10-15$$

Subtract 15 from 10 to get -5.

$$-8x+x^{2}=-5$$

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.

$$x^{2}-8x=-5$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-5+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-5+16$$

Add -5 to 16.

$$x^{2}-8x+16=11$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=11$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=\sqrt{11} x-4=-\sqrt{11}$$

Add 4 to both sides of the equation.

$$x=\sqrt{11}+4 x=4-\sqrt{11}$$