Solve 20n^2-98n=-48

$$20 n ^ { 2 } - 98 n = - 48$$

Solve for n

$$n = \frac{\sqrt{1441} + 49}{20} \approx 4.34802529$$

$$n=\frac{49-\sqrt{1441}}{20}\approx 0.55197471$$

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.

$$20n^{2}-98n=-48$$

Add 48 to both sides of the equation.

$$20n^{2}-98n-\left(-48\right)=-48-\left(-48\right)$$

Subtracting -48 from itself leaves 0.

$$20n^{2}-98n-\left(-48\right)=0$$

Subtract -48 from 0.

$$20n^{2}-98n+48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, -98 for b, and 48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-98\right)±\sqrt{\left(-98\right)^{2}-4\times 20\times 48}}{2\times 20}$$

Square -98.

$$n=\frac{-\left(-98\right)±\sqrt{9604-4\times 20\times 48}}{2\times 20}$$

Multiply -4 times 20.

$$n=\frac{-\left(-98\right)±\sqrt{9604-80\times 48}}{2\times 20}$$

Multiply -80 times 48.

$$n=\frac{-\left(-98\right)±\sqrt{9604-3840}}{2\times 20}$$

Add 9604 to -3840.

$$n=\frac{-\left(-98\right)±\sqrt{5764}}{2\times 20}$$

Take the square root of 5764.

$$n=\frac{-\left(-98\right)±2\sqrt{1441}}{2\times 20}$$

The opposite of -98 is 98.

$$n=\frac{98±2\sqrt{1441}}{2\times 20}$$

Multiply 2 times 20.

$$n=\frac{98±2\sqrt{1441}}{40}$$

Now solve the equation n=\frac{98±2\sqrt{1441}}{40} when ± is plus. Add 98 to 2\sqrt{1441}.

$$n=\frac{2\sqrt{1441}+98}{40}$$

Divide 98+2\sqrt{1441} by 40.

$$n=\frac{\sqrt{1441}+49}{20}$$

Now solve the equation n=\frac{98±2\sqrt{1441}}{40} when ± is minus. Subtract 2\sqrt{1441} from 98.

$$n=\frac{98-2\sqrt{1441}}{40}$$

Divide 98-2\sqrt{1441} by 40.

$$n=\frac{49-\sqrt{1441}}{20}$$

The equation is now solved.

$$n=\frac{\sqrt{1441}+49}{20} n=\frac{49-\sqrt{1441}}{20}$$

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.

$$20n^{2}-98n=-48$$

Divide both sides by 20.

$$\frac{20n^{2}-98n}{20}=-\frac{48}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$n^{2}+\left(-\frac{98}{20}\right)n=-\frac{48}{20}$$

Reduce the fraction \frac{-98}{20} to lowest terms by extracting and canceling out 2.

$$n^{2}-\frac{49}{10}n=-\frac{48}{20}$$

Reduce the fraction \frac{-48}{20} to lowest terms by extracting and canceling out 4.

$$n^{2}-\frac{49}{10}n=-\frac{12}{5}$$

Divide -\frac{49}{10}, the coefficient of the x term, by 2 to get -\frac{49}{20}. Then add the square of -\frac{49}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{49}{10}n+\left(-\frac{49}{20}\right)^{2}=-\frac{12}{5}+\left(-\frac{49}{20}\right)^{2}$$

Square -\frac{49}{20} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{49}{10}n+\frac{2401}{400}=-\frac{12}{5}+\frac{2401}{400}$$

Add -\frac{12}{5} to \frac{2401}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{49}{10}n+\frac{2401}{400}=\frac{1441}{400}$$

Factor n^{2}-\frac{49}{10}n+\frac{2401}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{49}{20}\right)^{2}=\frac{1441}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{49}{20}\right)^{2}}=\sqrt{\frac{1441}{400}}$$

Simplify.

$$n-\frac{49}{20}=\frac{\sqrt{1441}}{20} n-\frac{49}{20}=-\frac{\sqrt{1441}}{20}$$

Add \frac{49}{20} to both sides of the equation.

$$n=\frac{\sqrt{1441}+49}{20} n=\frac{49-\sqrt{1441}}{20}$$