Solve -3n^2+191n-2840=0

$$-3 { n }^{ 2 } +191n-2840=0$$

Solve for n

$$n=40$$

$$n = \frac{71}{3} = 23\frac{2}{3} \approx 23.666666667$$

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.

$$-3n^{2}+191n-2840=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 191 for b, and -2840 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-191±\sqrt{191^{2}-4\left(-3\right)\left(-2840\right)}}{2\left(-3\right)}$$

Square 191.

$$n=\frac{-191±\sqrt{36481-4\left(-3\right)\left(-2840\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$n=\frac{-191±\sqrt{36481+12\left(-2840\right)}}{2\left(-3\right)}$$

Multiply 12 times -2840.

$$n=\frac{-191±\sqrt{36481-34080}}{2\left(-3\right)}$$

Add 36481 to -34080.

$$n=\frac{-191±\sqrt{2401}}{2\left(-3\right)}$$

Take the square root of 2401.

$$n=\frac{-191±49}{2\left(-3\right)}$$

Multiply 2 times -3.

$$n=\frac{-191±49}{-6}$$

Now solve the equation n=\frac{-191±49}{-6} when ± is plus. Add -191 to 49.

$$n=-\frac{142}{-6}$$

Reduce the fraction \frac{-142}{-6} to lowest terms by extracting and canceling out 2.

$$n=\frac{71}{3}$$

Now solve the equation n=\frac{-191±49}{-6} when ± is minus. Subtract 49 from -191.

$$n=-\frac{240}{-6}$$

Divide -240 by -6.

$$n=40$$

The equation is now solved.

$$n=\frac{71}{3} n=40$$

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.

$$-3n^{2}+191n-2840=0$$

Add 2840 to both sides of the equation.

$$-3n^{2}+191n-2840-\left(-2840\right)=-\left(-2840\right)$$

Subtracting -2840 from itself leaves 0.

$$-3n^{2}+191n=-\left(-2840\right)$$

Subtract -2840 from 0.

$$-3n^{2}+191n=2840$$

Divide both sides by -3.

$$\frac{-3n^{2}+191n}{-3}=\frac{2840}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$n^{2}+\frac{191}{-3}n=\frac{2840}{-3}$$

Divide 191 by -3.

$$n^{2}-\frac{191}{3}n=\frac{2840}{-3}$$

Divide 2840 by -3.

$$n^{2}-\frac{191}{3}n=-\frac{2840}{3}$$

Divide -\frac{191}{3}, the coefficient of the x term, by 2 to get -\frac{191}{6}. Then add the square of -\frac{191}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{191}{3}n+\left(-\frac{191}{6}\right)^{2}=-\frac{2840}{3}+\left(-\frac{191}{6}\right)^{2}$$

Square -\frac{191}{6} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{191}{3}n+\frac{36481}{36}=-\frac{2840}{3}+\frac{36481}{36}$$

Add -\frac{2840}{3} to \frac{36481}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{191}{3}n+\frac{36481}{36}=\frac{2401}{36}$$

Factor n^{2}-\frac{191}{3}n+\frac{36481}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{191}{6}\right)^{2}=\frac{2401}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{191}{6}\right)^{2}}=\sqrt{\frac{2401}{36}}$$

Simplify.

$$n-\frac{191}{6}=\frac{49}{6} n-\frac{191}{6}=-\frac{49}{6}$$

Add \frac{191}{6} to both sides of the equation.

$$n=40 n=\frac{71}{3}$$