Solve n^2+6n=70+3n

$$n ^ { 2 } + 6 n = 70 + 3 n$$

Solve for n

$$n=-10$$

$$n=7$$

Steps Using Factoring

Subtract 70 from both sides.

$$n^{2}+6n-70=3n$$

Subtract 3n from both sides.

$$n^{2}+6n-70-3n=0$$

Combine 6n and -3n to get 3n.

$$n^{2}+3n-70=0$$

To solve the equation, factor n^{2}+3n-70 using formula n^{2}+\left(a+b\right)n+ab=\left(n+a\right)\left(n+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-70$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -70.

$$-1,70 -2,35 -5,14 -7,10$$

Calculate the sum for each pair.

$$-1+70=69 -2+35=33 -5+14=9 -7+10=3$$

The solution is the pair that gives sum 3.

$$a=-7 b=10$$

Rewrite factored expression \left(n+a\right)\left(n+b\right) using the obtained values.

$$\left(n-7\right)\left(n+10\right)$$

To find equation solutions, solve n-7=0 and n+10=0.

$$n=7 n=-10$$

Steps Using Factoring By Grouping

Subtract 70 from both sides.

$$n^{2}+6n-70=3n$$

Subtract 3n from both sides.

$$n^{2}+6n-70-3n=0$$

Combine 6n and -3n to get 3n.

$$n^{2}+3n-70=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as n^{2}+an+bn-70. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-70\right)=-70$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -70.

$$-1,70 -2,35 -5,14 -7,10$$

Calculate the sum for each pair.

$$-1+70=69 -2+35=33 -5+14=9 -7+10=3$$

The solution is the pair that gives sum 3.

$$a=-7 b=10$$

Rewrite n^{2}+3n-70 as \left(n^{2}-7n\right)+\left(10n-70\right).

$$\left(n^{2}-7n\right)+\left(10n-70\right)$$

Factor out n in the first and 10 in the second group.

$$n\left(n-7\right)+10\left(n-7\right)$$

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

$$\left(n-7\right)\left(n+10\right)$$

To find equation solutions, solve n-7=0 and n+10=0.

$$n=7 n=-10$$

Steps Using the Quadratic Formula

Subtract 70 from both sides.

$$n^{2}+6n-70=3n$$

Subtract 3n from both sides.

$$n^{2}+6n-70-3n=0$$

Combine 6n and -3n to get 3n.

$$n^{2}+3n-70=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}-4\left(-70\right)}}{2}$$

Square 3.

$$n=\frac{-3±\sqrt{9-4\left(-70\right)}}{2}$$

Multiply -4 times -70.

$$n=\frac{-3±\sqrt{9+280}}{2}$$

Add 9 to 280.

$$n=\frac{-3±\sqrt{289}}{2}$$

Take the square root of 289.

$$n=\frac{-3±17}{2}$$

Now solve the equation n=\frac{-3±17}{2} when ± is plus. Add -3 to 17.

$$n=\frac{14}{2}$$

Divide 14 by 2.

$$n=7$$

Now solve the equation n=\frac{-3±17}{2} when ± is minus. Subtract 17 from -3.

$$n=-\frac{20}{2}$$

Divide -20 by 2.

$$n=-10$$

The equation is now solved.

$$n=7 n=-10$$

Steps for Completing the Square

Subtract 3n from both sides.

$$n^{2}+6n-3n=70$$

Combine 6n and -3n to get 3n.

$$n^{2}+3n=70$$

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

$$n^{2}+3n+\left(\frac{3}{2}\right)^{2}=70+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+3n+\frac{9}{4}=70+\frac{9}{4}$$

Add 70 to \frac{9}{4}.

$$n^{2}+3n+\frac{9}{4}=\frac{289}{4}$$

Factor n^{2}+3n+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{3}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{3}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$n+\frac{3}{2}=\frac{17}{2} n+\frac{3}{2}=-\frac{17}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$n=7 n=-10$$