Solve x+52-2xdiv12=2+8x*2x+2*4

$$x+52-2x \div 12=2+8x \times 2x+2 \times 4$$

Solve for x

$$x = \frac{\sqrt{96793} + 5}{192} \approx 1.646436115$$

$$x=\frac{5-\sqrt{96793}}{192}\approx -1.594352782$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12.

$$12x+624-2x=24+96x\times 2x+24\times 4$$

Multiply x and x to get x^{2}.

$$12x+624-2x=24+96x^{2}\times 2+24\times 4$$

Combine 12x and -2x to get 10x.

$$10x+624=24+96x^{2}\times 2+24\times 4$$

Multiply 96 and 2 to get 192.

$$10x+624=24+192x^{2}+24\times 4$$

Multiply 24 and 4 to get 96.

$$10x+624=24+192x^{2}+96$$

Add 24 and 96 to get 120.

$$10x+624=120+192x^{2}$$

Subtract 120 from both sides.

$$10x+624-120=192x^{2}$$

Subtract 120 from 624 to get 504.

$$10x+504=192x^{2}$$

Subtract 192x^{2} from both sides.

$$10x+504-192x^{2}=0$$

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.

$$-192x^{2}+10x+504=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -192 for a, 10 for b, and 504 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-192\right)\times 504}}{2\left(-192\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-192\right)\times 504}}{2\left(-192\right)}$$

Multiply -4 times -192.

$$x=\frac{-10±\sqrt{100+768\times 504}}{2\left(-192\right)}$$

Multiply 768 times 504.

$$x=\frac{-10±\sqrt{100+387072}}{2\left(-192\right)}$$

Add 100 to 387072.

$$x=\frac{-10±\sqrt{387172}}{2\left(-192\right)}$$

Take the square root of 387172.

$$x=\frac{-10±2\sqrt{96793}}{2\left(-192\right)}$$

Multiply 2 times -192.

$$x=\frac{-10±2\sqrt{96793}}{-384}$$

Now solve the equation x=\frac{-10±2\sqrt{96793}}{-384} when ± is plus. Add -10 to 2\sqrt{96793}.

$$x=\frac{2\sqrt{96793}-10}{-384}$$

Divide -10+2\sqrt{96793} by -384.

$$x=\frac{5-\sqrt{96793}}{192}$$

Now solve the equation x=\frac{-10±2\sqrt{96793}}{-384} when ± is minus. Subtract 2\sqrt{96793} from -10.

$$x=\frac{-2\sqrt{96793}-10}{-384}$$

Divide -10-2\sqrt{96793} by -384.

$$x=\frac{\sqrt{96793}+5}{192}$$

The equation is now solved.

$$x=\frac{5-\sqrt{96793}}{192} x=\frac{\sqrt{96793}+5}{192}$$

Steps for Completing the Square

Multiply both sides of the equation by 12.

$$12x+624-2x=24+96x\times 2x+24\times 4$$

Multiply x and x to get x^{2}.

$$12x+624-2x=24+96x^{2}\times 2+24\times 4$$

Combine 12x and -2x to get 10x.

$$10x+624=24+96x^{2}\times 2+24\times 4$$

Multiply 96 and 2 to get 192.

$$10x+624=24+192x^{2}+24\times 4$$

Multiply 24 and 4 to get 96.

$$10x+624=24+192x^{2}+96$$

Add 24 and 96 to get 120.

$$10x+624=120+192x^{2}$$

Subtract 192x^{2} from both sides.

$$10x+624-192x^{2}=120$$

Subtract 624 from both sides.

$$10x-192x^{2}=120-624$$

Subtract 624 from 120 to get -504.

$$10x-192x^{2}=-504$$

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.

$$-192x^{2}+10x=-504$$

Divide both sides by -192.

$$\frac{-192x^{2}+10x}{-192}=-\frac{504}{-192}$$

Dividing by -192 undoes the multiplication by -192.

$$x^{2}+\frac{10}{-192}x=-\frac{504}{-192}$$

Reduce the fraction \frac{10}{-192} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{5}{96}x=-\frac{504}{-192}$$

Reduce the fraction \frac{-504}{-192} to lowest terms by extracting and canceling out 24.

$$x^{2}-\frac{5}{96}x=\frac{21}{8}$$

Divide -\frac{5}{96}, the coefficient of the x term, by 2 to get -\frac{5}{192}. Then add the square of -\frac{5}{192} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{96}x+\left(-\frac{5}{192}\right)^{2}=\frac{21}{8}+\left(-\frac{5}{192}\right)^{2}$$

Square -\frac{5}{192} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{96}x+\frac{25}{36864}=\frac{21}{8}+\frac{25}{36864}$$

Add \frac{21}{8} to \frac{25}{36864} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{96}x+\frac{25}{36864}=\frac{96793}{36864}$$

Factor x^{2}-\frac{5}{96}x+\frac{25}{36864}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{192}\right)^{2}=\frac{96793}{36864}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{192}\right)^{2}}=\sqrt{\frac{96793}{36864}}$$

Simplify.

$$x-\frac{5}{192}=\frac{\sqrt{96793}}{192} x-\frac{5}{192}=-\frac{\sqrt{96793}}{192}$$

Add \frac{5}{192} to both sides of the equation.

$$x=\frac{\sqrt{96793}+5}{192} x=\frac{5-\sqrt{96793}}{192}$$