Solve 20(n ^ 2 + 39n - 390) = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$20(n^{2}+39n-390)=0$$

Solve for n

$n = \frac{\sqrt{3081} - 39}{2} \approx 8.253378173$
$n=\frac{-\sqrt{3081}-39}{2}\approx -47.253378173$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Divide both sides by $20$. Zero divided by any non-zero number gives zero.

$$n^{2}+39n-390=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $39$ for $b$, and $-390$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{-39±\sqrt{39^{2}-4\left(-390\right)}}{2}$$

Square $39$.

$$n=\frac{-39±\sqrt{1521-4\left(-390\right)}}{2}$$

Multiply $-4$ times $-390$.

$$n=\frac{-39±\sqrt{1521+1560}}{2}$$

Add $1521$ to $1560$.

$$n=\frac{-39±\sqrt{3081}}{2}$$

Now solve the equation $n=\frac{-39±\sqrt{3081}}{2}$ when $±$ is plus. Add $-39$ to $\sqrt{3081}$.

$$n=\frac{\sqrt{3081}-39}{2}$$

Now solve the equation $n=\frac{-39±\sqrt{3081}}{2}$ when $±$ is minus. Subtract $\sqrt{3081}$ from $-39$.

$$n=\frac{-\sqrt{3081}-39}{2}$$

The equation is now solved.

$$n=\frac{\sqrt{3081}-39}{2}$$ $$n=\frac{-\sqrt{3081}-39}{2}$$

Steps for Completing the Square

Divide both sides by $20$. Zero divided by any non-zero number gives zero.

$$n^{2}+39n-390=0$$

Add $390$ to both sides. Anything plus zero gives itself.

$$n^{2}+39n=390$$

Divide $39$, the coefficient of the $x$ term, by $2$ to get $\frac{39}{2}$. Then add the square of $\frac{39}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+39n+\left(\frac{39}{2}\right)^{2}=390+\left(\frac{39}{2}\right)^{2}$$

Square $\frac{39}{2}$ by squaring both the numerator and the denominator of the fraction.

$$n^{2}+39n+\frac{1521}{4}=390+\frac{1521}{4}$$

Add $390$ to $\frac{1521}{4}$.

$$n^{2}+39n+\frac{1521}{4}=\frac{3081}{4}$$

Factor $n^{2}+39n+\frac{1521}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(n+\frac{39}{2}\right)^{2}=\frac{3081}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{39}{2}\right)^{2}}=\sqrt{\frac{3081}{4}}$$

Simplify.

$$n+\frac{39}{2}=\frac{\sqrt{3081}}{2}$$ $$n+\frac{39}{2}=-\frac{\sqrt{3081}}{2}$$

Subtract $\frac{39}{2}$ from both sides of the equation.

$$n=\frac{\sqrt{3081}-39}{2}$$ $$n=\frac{-\sqrt{3081}-39}{2}$$