How to solve (n(n-3))/(2)=77

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (n(n-3))/(2)=77 | Equatix

Question

$$\frac{ n(n-3) }{ 2 } =77$$

Solve for n

$n=-11$
$n=14$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides by $2$.

$$n\left(n-3\right)=77\times 2$$

Use the distributive property to multiply $n$ by $n-3$.

$$n^{2}-3n=77\times 2$$

Multiply $77$ and $2$ to get $154$.

$$n^{2}-3n=154$$

Subtract $154$ from both sides.

$$n^{2}-3n-154=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-3$ for $b$, and $-154$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-154\right)}}{2}$$

Square $-3$.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\left(-154\right)}}{2}$$

Multiply $-4$ times $-154$.

$$n=\frac{-\left(-3\right)±\sqrt{9+616}}{2}$$

Add $9$ to $616$.

$$n=\frac{-\left(-3\right)±\sqrt{625}}{2}$$

Take the square root of $625$.

$$n=\frac{-\left(-3\right)±25}{2}$$

The opposite of $-3$ is $3$.

$$n=\frac{3±25}{2}$$

Now solve the equation $n=\frac{3±25}{2}$ when $±$ is plus. Add $3$ to $25$.

$$n=\frac{28}{2}$$

Divide $28$ by $2$.

$$n=14$$

Now solve the equation $n=\frac{3±25}{2}$ when $±$ is minus. Subtract $25$ from $3$.

$$n=-\frac{22}{2}$$

Divide $-22$ by $2$.

$$n=-11$$

The equation is now solved.

$$n=14$$ $$n=-11$$

Steps for Completing the Square

Multiply both sides by $2$.

$$n\left(n-3\right)=77\times 2$$

Use the distributive property to multiply $n$ by $n-3$.

$$n^{2}-3n=77\times 2$$

Multiply $77$ and $2$ to get $154$.

$$n^{2}-3n=154$$

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.

$$n^{2}-3n+\left(-\frac{3}{2}\right)^{2}=154+\left(-\frac{3}{2}\right)^{2}$$

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

$$n^{2}-3n+\frac{9}{4}=154+\frac{9}{4}$$

Add $154$ to $\frac{9}{4}$.

$$n^{2}-3n+\frac{9}{4}=\frac{625}{4}$$

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

$$\left(n-\frac{3}{2}\right)^{2}=\frac{625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{3}{2}\right)^{2}}=\sqrt{\frac{625}{4}}$$

Simplify.

$$n-\frac{3}{2}=\frac{25}{2}$$ $$n-\frac{3}{2}=-\frac{25}{2}$$

Add $\frac{3}{2}$ to both sides of the equation.

$$n=14$$ $$n=-11$$

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Question

Solve for n

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square