Solve [8x(2-x)-(4x^2-7)]/(2-x)^2=0

$$[ 8 x ( 2 - x ) - ( 4 x ^ { 2 } - 7 ) ] / ( 2 - x ) ^ { 2 } = 0$$

Solve for x

$$x=\frac{\sqrt{37}}{6}+\frac{2}{3}\approx 1.680460422$$

$$x=-\frac{\sqrt{37}}{6}+\frac{2}{3}\approx -0.347127088$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

$$8x\left(2-x\right)-\left(4x^{2}-7\right)=0$$

Use the distributive property to multiply 8x by 2-x.

$$16x-8x^{2}-\left(4x^{2}-7\right)=0$$

To find the opposite of 4x^{2}-7, find the opposite of each term.

$$16x-8x^{2}-4x^{2}+7=0$$

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$16x-12x^{2}+7=0$$

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.

$$-12x^{2}+16x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, 16 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-12\right)\times 7}}{2\left(-12\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-12\right)\times 7}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-16±\sqrt{256+48\times 7}}{2\left(-12\right)}$$

Multiply 48 times 7.

$$x=\frac{-16±\sqrt{256+336}}{2\left(-12\right)}$$

Add 256 to 336.

$$x=\frac{-16±\sqrt{592}}{2\left(-12\right)}$$

Take the square root of 592.

$$x=\frac{-16±4\sqrt{37}}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-16±4\sqrt{37}}{-24}$$

Now solve the equation x=\frac{-16±4\sqrt{37}}{-24} when ± is plus. Add -16 to 4\sqrt{37}.

$$x=\frac{4\sqrt{37}-16}{-24}$$

Divide -16+4\sqrt{37} by -24.

$$x=-\frac{\sqrt{37}}{6}+\frac{2}{3}$$

Now solve the equation x=\frac{-16±4\sqrt{37}}{-24} when ± is minus. Subtract 4\sqrt{37} from -16.

$$x=\frac{-4\sqrt{37}-16}{-24}$$

Divide -16-4\sqrt{37} by -24.

$$x=\frac{\sqrt{37}}{6}+\frac{2}{3}$$

The equation is now solved.

$$x=-\frac{\sqrt{37}}{6}+\frac{2}{3} x=\frac{\sqrt{37}}{6}+\frac{2}{3}$$

Steps for Completing the Square

Variable x cannot be equal to 2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)^{2}.

$$8x\left(2-x\right)-\left(4x^{2}-7\right)=0$$

Use the distributive property to multiply 8x by 2-x.

$$16x-8x^{2}-\left(4x^{2}-7\right)=0$$

To find the opposite of 4x^{2}-7, find the opposite of each term.

$$16x-8x^{2}-4x^{2}+7=0$$

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$16x-12x^{2}+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$16x-12x^{2}=-7$$

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.

$$-12x^{2}+16x=-7$$

Divide both sides by -12.

$$\frac{-12x^{2}+16x}{-12}=-\frac{7}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{16}{-12}x=-\frac{7}{-12}$$

Reduce the fraction \frac{16}{-12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{4}{3}x=-\frac{7}{-12}$$

Divide -7 by -12.

$$x^{2}-\frac{4}{3}x=\frac{7}{12}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{7}{12}+\left(-\frac{2}{3}\right)^{2}$$

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

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{7}{12}+\frac{4}{9}$$

Add \frac{7}{12} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{37}{36}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$x-\frac{2}{3}=\frac{\sqrt{37}}{6} x-\frac{2}{3}=-\frac{\sqrt{37}}{6}$$

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

$$x=\frac{\sqrt{37}}{6}+\frac{2}{3} x=-\frac{\sqrt{37}}{6}+\frac{2}{3}$$