Solve (7-2x)(x-2)=1.12

$$(7-2x)(x-2)=1.12$$

Solve for x

$$x=2.8$$

$$x=2.7$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 7-2x by x-2 and combine like terms.

$$11x-14-2x^{2}=1.12$$

Subtract 1.12 from both sides.

$$11x-14-2x^{2}-1.12=0$$

Subtract 1.12 from -14 to get -15.12.

$$11x-15.12-2x^{2}=0$$

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.

$$-2x^{2}+11x-15.12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 11 for b, and -15.12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\left(-15.12\right)}}{2\left(-2\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-2\right)\left(-15.12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-11±\sqrt{121+8\left(-15.12\right)}}{2\left(-2\right)}$$

Multiply 8 times -15.12.

$$x=\frac{-11±\sqrt{121-120.96}}{2\left(-2\right)}$$

Add 121 to -120.96.

$$x=\frac{-11±\sqrt{0.04}}{2\left(-2\right)}$$

Take the square root of 0.04.

$$x=\frac{-11±\frac{1}{5}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-11±\frac{1}{5}}{-4}$$

Now solve the equation x=\frac{-11±\frac{1}{5}}{-4} when ± is plus. Add -11 to \frac{1}{5}.

$$x=-\frac{\frac{54}{5}}{-4}$$

Divide -\frac{54}{5} by -4.

$$x=\frac{27}{10}$$

Now solve the equation x=\frac{-11±\frac{1}{5}}{-4} when ± is minus. Subtract \frac{1}{5} from -11.

$$x=-\frac{\frac{56}{5}}{-4}$$

Divide -\frac{56}{5} by -4.

$$x=\frac{14}{5}$$

The equation is now solved.

$$x=\frac{27}{10} x=\frac{14}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 7-2x by x-2 and combine like terms.

$$11x-14-2x^{2}=1.12$$

Add 14 to both sides.

$$11x-2x^{2}=1.12+14$$

Add 1.12 and 14 to get 15.12.

$$11x-2x^{2}=15.12$$

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.

$$-2x^{2}+11x=15.12$$

Divide both sides by -2.

$$\frac{-2x^{2}+11x}{-2}=\frac{15.12}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{11}{-2}x=\frac{15.12}{-2}$$

Divide 11 by -2.

$$x^{2}-\frac{11}{2}x=\frac{15.12}{-2}$$

Divide 15.12 by -2.

$$x^{2}-\frac{11}{2}x=-7.56$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-7.56+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-7.56+\frac{121}{16}$$

Add -7.56 to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{1}{400}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{1}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{1}{400}}$$

Simplify.

$$x-\frac{11}{4}=\frac{1}{20} x-\frac{11}{4}=-\frac{1}{20}$$

Add \frac{11}{4} to both sides of the equation.

$$x=\frac{14}{5} x=\frac{27}{10}$$