Solve 3x^2+2-(5x-50)^2=0

$$3 x ^ { 2 } + 2 - ( 5 x - 50 ) ^ { 2 } = 0$$

Solve for x

$$x = \frac{\sqrt{1886} + 125}{11} \approx 15.311645591$$

$$x = \frac{125 - \sqrt{1886}}{11} \approx 7.415627136$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-50\right)^{2}.

$$3x^{2}+2-\left(25x^{2}-500x+2500\right)=0$$

To find the opposite of 25x^{2}-500x+2500, find the opposite of each term.

$$3x^{2}+2-25x^{2}+500x-2500=0$$

Combine 3x^{2} and -25x^{2} to get -22x^{2}.

$$-22x^{2}+2+500x-2500=0$$

Subtract 2500 from 2 to get -2498.

$$-22x^{2}-2498+500x=0$$

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.

$$-22x^{2}+500x-2498=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -22 for a, 500 for b, and -2498 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-500±\sqrt{500^{2}-4\left(-22\right)\left(-2498\right)}}{2\left(-22\right)}$$

Square 500.

$$x=\frac{-500±\sqrt{250000-4\left(-22\right)\left(-2498\right)}}{2\left(-22\right)}$$

Multiply -4 times -22.

$$x=\frac{-500±\sqrt{250000+88\left(-2498\right)}}{2\left(-22\right)}$$

Multiply 88 times -2498.

$$x=\frac{-500±\sqrt{250000-219824}}{2\left(-22\right)}$$

Add 250000 to -219824.

$$x=\frac{-500±\sqrt{30176}}{2\left(-22\right)}$$

Take the square root of 30176.

$$x=\frac{-500±4\sqrt{1886}}{2\left(-22\right)}$$

Multiply 2 times -22.

$$x=\frac{-500±4\sqrt{1886}}{-44}$$

Now solve the equation x=\frac{-500±4\sqrt{1886}}{-44} when ± is plus. Add -500 to 4\sqrt{1886}.

$$x=\frac{4\sqrt{1886}-500}{-44}$$

Divide -500+4\sqrt{1886} by -44.

$$x=\frac{125-\sqrt{1886}}{11}$$

Now solve the equation x=\frac{-500±4\sqrt{1886}}{-44} when ± is minus. Subtract 4\sqrt{1886} from -500.

$$x=\frac{-4\sqrt{1886}-500}{-44}$$

Divide -500-4\sqrt{1886} by -44.

$$x=\frac{\sqrt{1886}+125}{11}$$

The equation is now solved.

$$x=\frac{125-\sqrt{1886}}{11} x=\frac{\sqrt{1886}+125}{11}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-50\right)^{2}.

$$3x^{2}+2-\left(25x^{2}-500x+2500\right)=0$$

To find the opposite of 25x^{2}-500x+2500, find the opposite of each term.

$$3x^{2}+2-25x^{2}+500x-2500=0$$

Combine 3x^{2} and -25x^{2} to get -22x^{2}.

$$-22x^{2}+2+500x-2500=0$$

Subtract 2500 from 2 to get -2498.

$$-22x^{2}-2498+500x=0$$

Add 2498 to both sides. Anything plus zero gives itself.

$$-22x^{2}+500x=2498$$

Divide both sides by -22.

$$\frac{-22x^{2}+500x}{-22}=\frac{2498}{-22}$$

Dividing by -22 undoes the multiplication by -22.

$$x^{2}+\frac{500}{-22}x=\frac{2498}{-22}$$

Reduce the fraction \frac{500}{-22} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{250}{11}x=\frac{2498}{-22}$$

Reduce the fraction \frac{2498}{-22} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{250}{11}x=-\frac{1249}{11}$$

Divide -\frac{250}{11}, the coefficient of the x term, by 2 to get -\frac{125}{11}. Then add the square of -\frac{125}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{250}{11}x+\left(-\frac{125}{11}\right)^{2}=-\frac{1249}{11}+\left(-\frac{125}{11}\right)^{2}$$

Square -\frac{125}{11} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{250}{11}x+\frac{15625}{121}=-\frac{1249}{11}+\frac{15625}{121}$$

Add -\frac{1249}{11} to \frac{15625}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{250}{11}x+\frac{15625}{121}=\frac{1886}{121}$$

Factor x^{2}-\frac{250}{11}x+\frac{15625}{121}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{125}{11}\right)^{2}=\frac{1886}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{125}{11}\right)^{2}}=\sqrt{\frac{1886}{121}}$$

Simplify.

$$x-\frac{125}{11}=\frac{\sqrt{1886}}{11} x-\frac{125}{11}=-\frac{\sqrt{1886}}{11}$$

Add \frac{125}{11} to both sides of the equation.

$$x=\frac{\sqrt{1886}+125}{11} x=\frac{125-\sqrt{1886}}{11}$$