Solve (20-5x)(6-x)=12.5*6

$$( 20 - 5 x ) ( 6 - x ) = 12.5 \times 6$$

Solve for x

$$x=1$$

$$x=9$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20-5x by 6-x and combine like terms.

$$120-50x+5x^{2}=12.5\times 6$$

Multiply 12.5 and 6 to get 75.

$$120-50x+5x^{2}=75$$

Subtract 75 from both sides.

$$120-50x+5x^{2}-75=0$$

Subtract 75 from 120 to get 45.

$$45-50x+5x^{2}=0$$

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.

$$5x^{2}-50x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -50 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 5\times 45}}{2\times 5}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 5\times 45}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-50\right)±\sqrt{2500-20\times 45}}{2\times 5}$$

Multiply -20 times 45.

$$x=\frac{-\left(-50\right)±\sqrt{2500-900}}{2\times 5}$$

Add 2500 to -900.

$$x=\frac{-\left(-50\right)±\sqrt{1600}}{2\times 5}$$

Take the square root of 1600.

$$x=\frac{-\left(-50\right)±40}{2\times 5}$$

The opposite of -50 is 50.

$$x=\frac{50±40}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{50±40}{10}$$

Now solve the equation x=\frac{50±40}{10} when ± is plus. Add 50 to 40.

$$x=\frac{90}{10}$$

Divide 90 by 10.

$$x=9$$

Now solve the equation x=\frac{50±40}{10} when ± is minus. Subtract 40 from 50.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

The equation is now solved.

$$x=9 x=1$$

Steps for Completing the Square

Use the distributive property to multiply 20-5x by 6-x and combine like terms.

$$120-50x+5x^{2}=12.5\times 6$$

Multiply 12.5 and 6 to get 75.

$$120-50x+5x^{2}=75$$

Subtract 120 from both sides.

$$-50x+5x^{2}=75-120$$

Subtract 120 from 75 to get -45.

$$-50x+5x^{2}=-45$$

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.

$$5x^{2}-50x=-45$$

Divide both sides by 5.

$$\frac{5x^{2}-50x}{5}=-\frac{45}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{50}{5}\right)x=-\frac{45}{5}$$

Divide -50 by 5.

$$x^{2}-10x=-\frac{45}{5}$$

Divide -45 by 5.

$$x^{2}-10x=-9$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-9+25$$

Add -9 to 25.

$$x^{2}-10x+25=16$$

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}.

$$\left(x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-5=4 x-5=-4$$

Add 5 to both sides of the equation.

$$x=9 x=1$$