Solve 10x^2+2x-25=100

$$10 x ^ { 2 } + 2 x - 25 = 100$$

Solve for x

$$x = \frac{3 \sqrt{139} - 1}{10} \approx 3.436947837$$

$$x=\frac{-3\sqrt{139}-1}{10}\approx -3.636947837$$

Steps Using the Quadratic Formula

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.

$$10x^{2}+2x-25=100$$

Subtract 100 from both sides of the equation.

$$10x^{2}+2x-25-100=100-100$$

Subtracting 100 from itself leaves 0.

$$10x^{2}+2x-25-100=0$$

Subtract 100 from -25.

$$10x^{2}+2x-125=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 2 for b, and -125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 10\left(-125\right)}}{2\times 10}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 10\left(-125\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-2±\sqrt{4-40\left(-125\right)}}{2\times 10}$$

Multiply -40 times -125.

$$x=\frac{-2±\sqrt{4+5000}}{2\times 10}$$

Add 4 to 5000.

$$x=\frac{-2±\sqrt{5004}}{2\times 10}$$

Take the square root of 5004.

$$x=\frac{-2±6\sqrt{139}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-2±6\sqrt{139}}{20}$$

Now solve the equation x=\frac{-2±6\sqrt{139}}{20} when ± is plus. Add -2 to 6\sqrt{139}.

$$x=\frac{6\sqrt{139}-2}{20}$$

Divide -2+6\sqrt{139} by 20.

$$x=\frac{3\sqrt{139}-1}{10}$$

Now solve the equation x=\frac{-2±6\sqrt{139}}{20} when ± is minus. Subtract 6\sqrt{139} from -2.

$$x=\frac{-6\sqrt{139}-2}{20}$$

Divide -2-6\sqrt{139} by 20.

$$x=\frac{-3\sqrt{139}-1}{10}$$

The equation is now solved.

$$x=\frac{3\sqrt{139}-1}{10} x=\frac{-3\sqrt{139}-1}{10}$$

Steps for Completing the Square

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.

$$10x^{2}+2x-25=100$$

Add 25 to both sides of the equation.

$$10x^{2}+2x-25-\left(-25\right)=100-\left(-25\right)$$

Subtracting -25 from itself leaves 0.

$$10x^{2}+2x=100-\left(-25\right)$$

Subtract -25 from 100.

$$10x^{2}+2x=125$$

Divide both sides by 10.

$$\frac{10x^{2}+2x}{10}=\frac{125}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{2}{10}x=\frac{125}{10}$$

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

$$x^{2}+\frac{1}{5}x=\frac{125}{10}$$

Reduce the fraction \frac{125}{10} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{1}{5}x=\frac{25}{2}$$

Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{25}{2}+\left(\frac{1}{10}\right)^{2}$$

Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{25}{2}+\frac{1}{100}$$

Add \frac{25}{2} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{1251}{100}$$

Factor x^{2}+\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{1251}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{1251}{100}}$$

Simplify.

$$x+\frac{1}{10}=\frac{3\sqrt{139}}{10} x+\frac{1}{10}=-\frac{3\sqrt{139}}{10}$$

Subtract \frac{1}{10} from both sides of the equation.

$$x=\frac{3\sqrt{139}-1}{10} x=\frac{-3\sqrt{139}-1}{10}$$