Solve (x+3)(3-0.5x)=10

$$( x + 3 ) ( 3 - 0.5 x ) = 10$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x+3 by 3-0.5x and combine like terms.

$$1.5x-0.5x^{2}+9=10$$

Subtract 10 from both sides.

$$1.5x-0.5x^{2}+9-10=0$$

Subtract 10 from 9 to get -1.

$$1.5x-0.5x^{2}-1=0$$

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.

$$-0.5x^{2}+1.5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.5 for a, 1.5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1.5±\sqrt{1.5^{2}-4\left(-0.5\right)\left(-1\right)}}{2\left(-0.5\right)}$$

Square 1.5 by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-1.5±\sqrt{2.25-4\left(-0.5\right)\left(-1\right)}}{2\left(-0.5\right)}$$

Multiply -4 times -0.5.

$$x=\frac{-1.5±\sqrt{2.25+2\left(-1\right)}}{2\left(-0.5\right)}$$

Multiply 2 times -1.

$$x=\frac{-1.5±\sqrt{2.25-2}}{2\left(-0.5\right)}$$

Add 2.25 to -2.

$$x=\frac{-1.5±\sqrt{0.25}}{2\left(-0.5\right)}$$

Take the square root of 0.25.

$$x=\frac{-1.5±\frac{1}{2}}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{-1.5±\frac{1}{2}}{-1}$$

Now solve the equation x=\frac{-1.5±\frac{1}{2}}{-1} when ± is plus. Add -1.5 to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -1 by -1.

$$x=1$$

Now solve the equation x=\frac{-1.5±\frac{1}{2}}{-1} when ± is minus. Subtract \frac{1}{2} from -1.5 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -2 by -1.

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

Use the distributive property to multiply x+3 by 3-0.5x and combine like terms.

$$1.5x-0.5x^{2}+9=10$$

Subtract 9 from both sides.

$$1.5x-0.5x^{2}=10-9$$

Subtract 9 from 10 to get 1.

$$1.5x-0.5x^{2}=1$$

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.

$$-0.5x^{2}+1.5x=1$$

Multiply both sides by -2.

$$\frac{-0.5x^{2}+1.5x}{-0.5}=\frac{1}{-0.5}$$

Dividing by -0.5 undoes the multiplication by -0.5.

$$x^{2}+\frac{1.5}{-0.5}x=\frac{1}{-0.5}$$

Divide 1.5 by -0.5 by multiplying 1.5 by the reciprocal of -0.5.

$$x^{2}-3x=\frac{1}{-0.5}$$

Divide 1 by -0.5 by multiplying 1 by the reciprocal of -0.5.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=0.25$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=0.25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{0.25}$$

Simplify.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$