Solve (4+x)*(5-0.5x)=24

$$( 4 + x ) \times ( 5 - 0.5 x ) = 24$$

Solve for x

$$x=2$$

$$x=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 4+x by 5-0.5x and combine like terms.

$$20+3x-0.5x^{2}=24$$

Subtract 24 from both sides.

$$20+3x-0.5x^{2}-24=0$$

Subtract 24 from 20 to get -4.

$$-4+3x-0.5x^{2}=0$$

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.

$$-0.5x^{2}+3x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.5 for a, 3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-0.5\right)\left(-4\right)}}{2\left(-0.5\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-0.5\right)\left(-4\right)}}{2\left(-0.5\right)}$$

Multiply -4 times -0.5.

$$x=\frac{-3±\sqrt{9+2\left(-4\right)}}{2\left(-0.5\right)}$$

Multiply 2 times -4.

$$x=\frac{-3±\sqrt{9-8}}{2\left(-0.5\right)}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2\left(-0.5\right)}$$

Take the square root of 1.

$$x=\frac{-3±1}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{-3±1}{-1}$$

Now solve the equation x=\frac{-3±1}{-1} when ± is plus. Add -3 to 1.

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

Divide -2 by -1.

$$x=2$$

Now solve the equation x=\frac{-3±1}{-1} when ± is minus. Subtract 1 from -3.

$$x=-\frac{4}{-1}$$

Divide -4 by -1.

$$x=4$$

The equation is now solved.

$$x=2 x=4$$

Steps for Completing the Square

Use the distributive property to multiply 4+x by 5-0.5x and combine like terms.

$$20+3x-0.5x^{2}=24$$

Subtract 20 from both sides.

$$3x-0.5x^{2}=24-20$$

Subtract 20 from 24 to get 4.

$$3x-0.5x^{2}=4$$

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.

$$-0.5x^{2}+3x=4$$

Multiply both sides by -2.

$$\frac{-0.5x^{2}+3x}{-0.5}=\frac{4}{-0.5}$$

Dividing by -0.5 undoes the multiplication by -0.5.

$$x^{2}+\frac{3}{-0.5}x=\frac{4}{-0.5}$$

Divide 3 by -0.5 by multiplying 3 by the reciprocal of -0.5.

$$x^{2}-6x=\frac{4}{-0.5}$$

Divide 4 by -0.5 by multiplying 4 by the reciprocal of -0.5.

$$x^{2}-6x=-8$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-8+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-8+9$$

Add -8 to 9.

$$x^{2}-6x+9=1$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{1}$$

Simplify.

$$x-3=1 x-3=-1$$

Add 3 to both sides of the equation.

$$x=4 x=2$$