Solve 3n^2-28n=-49

$$3 n ^ { 2 } - 28 n = - 49$$

Solve for n

$$n = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$n=7$$

Steps Using Factoring By Grouping

Add 49 to both sides.

$$3n^{2}-28n+49=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3n^{2}+an+bn+49. To find a and b, set up a system to be solved.

$$a+b=-28 ab=3\times 49=147$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 147.

$$-1,-147 -3,-49 -7,-21$$

Calculate the sum for each pair.

$$-1-147=-148 -3-49=-52 -7-21=-28$$

The solution is the pair that gives sum -28.

$$a=-21 b=-7$$

Rewrite 3n^{2}-28n+49 as \left(3n^{2}-21n\right)+\left(-7n+49\right).

$$\left(3n^{2}-21n\right)+\left(-7n+49\right)$$

Factor out 3n in the first and -7 in the second group.

$$3n\left(n-7\right)-7\left(n-7\right)$$

Factor out common term n-7 by using distributive property.

$$\left(n-7\right)\left(3n-7\right)$$

To find equation solutions, solve n-7=0 and 3n-7=0.

$$n=7 n=\frac{7}{3}$$

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.

$$3n^{2}-28n=-49$$

Add 49 to both sides of the equation.

$$3n^{2}-28n-\left(-49\right)=-49-\left(-49\right)$$

Subtracting -49 from itself leaves 0.

$$3n^{2}-28n-\left(-49\right)=0$$

Subtract -49 from 0.

$$3n^{2}-28n+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -28 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 3\times 49}}{2\times 3}$$

Square -28.

$$n=\frac{-\left(-28\right)±\sqrt{784-4\times 3\times 49}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-\left(-28\right)±\sqrt{784-12\times 49}}{2\times 3}$$

Multiply -12 times 49.

$$n=\frac{-\left(-28\right)±\sqrt{784-588}}{2\times 3}$$

Add 784 to -588.

$$n=\frac{-\left(-28\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$n=\frac{-\left(-28\right)±14}{2\times 3}$$

The opposite of -28 is 28.

$$n=\frac{28±14}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{28±14}{6}$$

Now solve the equation n=\frac{28±14}{6} when ± is plus. Add 28 to 14.

$$n=\frac{42}{6}$$

Divide 42 by 6.

$$n=7$$

Now solve the equation n=\frac{28±14}{6} when ± is minus. Subtract 14 from 28.

$$n=\frac{14}{6}$$

Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

$$n=\frac{7}{3}$$

The equation is now solved.

$$n=7 n=\frac{7}{3}$$

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.

$$3n^{2}-28n=-49$$

Divide both sides by 3.

$$\frac{3n^{2}-28n}{3}=-\frac{49}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$n^{2}-\frac{28}{3}n=-\frac{49}{3}$$

Divide -\frac{28}{3}, the coefficient of the x term, by 2 to get -\frac{14}{3}. Then add the square of -\frac{14}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{28}{3}n+\left(-\frac{14}{3}\right)^{2}=-\frac{49}{3}+\left(-\frac{14}{3}\right)^{2}$$

Square -\frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{28}{3}n+\frac{196}{9}=-\frac{49}{3}+\frac{196}{9}$$

Add -\frac{49}{3} to \frac{196}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{28}{3}n+\frac{196}{9}=\frac{49}{9}$$

Factor n^{2}-\frac{28}{3}n+\frac{196}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{14}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{14}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$n-\frac{14}{3}=\frac{7}{3} n-\frac{14}{3}=-\frac{7}{3}$$

Add \frac{14}{3} to both sides of the equation.

$$n=7 n=\frac{7}{3}$$