Solve (10-x)(8+2x)=90

$$(10-x)(8+2x)=90$$

Solve for x

$$x=1$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 10-x by 8+2x and combine like terms.

$$80+12x-2x^{2}=90$$

Subtract 90 from both sides.

$$80+12x-2x^{2}-90=0$$

Subtract 90 from 80 to get -10.

$$-10+12x-2x^{2}=0$$

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.

$$-2x^{2}+12x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-10\right)}}{2\left(-2\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-2\right)\left(-10\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-12±\sqrt{144+8\left(-10\right)}}{2\left(-2\right)}$$

Multiply 8 times -10.

$$x=\frac{-12±\sqrt{144-80}}{2\left(-2\right)}$$

Add 144 to -80.

$$x=\frac{-12±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

$$x=\frac{-12±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-12±8}{-4}$$

Now solve the equation x=\frac{-12±8}{-4} when ± is plus. Add -12 to 8.

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

Divide -4 by -4.

$$x=1$$

Now solve the equation x=\frac{-12±8}{-4} when ± is minus. Subtract 8 from -12.

$$x=-\frac{20}{-4}$$

Divide -20 by -4.

$$x=5$$

The equation is now solved.

$$x=1 x=5$$

Steps for Completing the Square

Use the distributive property to multiply 10-x by 8+2x and combine like terms.

$$80+12x-2x^{2}=90$$

Subtract 80 from both sides.

$$12x-2x^{2}=90-80$$

Subtract 80 from 90 to get 10.

$$12x-2x^{2}=10$$

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.

$$-2x^{2}+12x=10$$

Divide both sides by -2.

$$\frac{-2x^{2}+12x}{-2}=\frac{10}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{12}{-2}x=\frac{10}{-2}$$

Divide 12 by -2.

$$x^{2}-6x=\frac{10}{-2}$$

Divide 10 by -2.

$$x^{2}-6x=-5$$

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.

$$x^{2}-6x+\left(-3\right)^{2}=-5+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-5+9$$

Add -5 to 9.

$$x^{2}-6x+9=4$$

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}.

$$\left(x-3\right)^{2}=4$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-3=2 x-3=-2$$

Add 3 to both sides of the equation.

$$x=5 x=1$$