Solve 0 = 2n ^ 2 + 23n + 231 | 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

$$0=2n^{2}+23n+231$$

Solve for n

$n=\frac{-23+\sqrt{1319}i}{4}\approx -5.75+9.079509899i$
$n=\frac{-\sqrt{1319}i-23}{4}\approx -5.75-9.079509899i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}+23n+231=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $23$ for $b$, and $231$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{-23±\sqrt{23^{2}-4\times 2\times 231}}{2\times 2}$$

Square $23$.

$$n=\frac{-23±\sqrt{529-4\times 2\times 231}}{2\times 2}$$

Multiply $-4$ times $2$.

$$n=\frac{-23±\sqrt{529-8\times 231}}{2\times 2}$$

Multiply $-8$ times $231$.

$$n=\frac{-23±\sqrt{529-1848}}{2\times 2}$$

Add $529$ to $-1848$.

$$n=\frac{-23±\sqrt{-1319}}{2\times 2}$$

Take the square root of $-1319$.

$$n=\frac{-23±\sqrt{1319}i}{2\times 2}$$

Multiply $2$ times $2$.

$$n=\frac{-23±\sqrt{1319}i}{4}$$

Now solve the equation $n=\frac{-23±\sqrt{1319}i}{4}$ when $±$ is plus. Add $-23$ to $i\sqrt{1319}$.

$$n=\frac{-23+\sqrt{1319}i}{4}$$

Now solve the equation $n=\frac{-23±\sqrt{1319}i}{4}$ when $±$ is minus. Subtract $i\sqrt{1319}$ from $-23$.

$$n=\frac{-\sqrt{1319}i-23}{4}$$

The equation is now solved.

$$n=\frac{-23+\sqrt{1319}i}{4}$$ $$n=\frac{-\sqrt{1319}i-23}{4}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$2n^{2}+23n+231=0$$

Subtract $231$ from both sides. Anything subtracted from zero gives its negation.

$$2n^{2}+23n=-231$$

Divide both sides by $2$.

$$\frac{2n^{2}+23n}{2}=-\frac{231}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$n^{2}+\frac{23}{2}n=-\frac{231}{2}$$

Divide $\frac{23}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{23}{4}$. Then add the square of $\frac{23}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{23}{2}n+\left(\frac{23}{4}\right)^{2}=-\frac{231}{2}+\left(\frac{23}{4}\right)^{2}$$

Square $\frac{23}{4}$ by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{23}{2}n+\frac{529}{16}=-\frac{231}{2}+\frac{529}{16}$$

Add $-\frac{231}{2}$ to $\frac{529}{16}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{23}{2}n+\frac{529}{16}=-\frac{1319}{16}$$

Factor $n^{2}+\frac{23}{2}n+\frac{529}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(n+\frac{23}{4}\right)^{2}=-\frac{1319}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{23}{4}\right)^{2}}=\sqrt{-\frac{1319}{16}}$$

Simplify.

$$n+\frac{23}{4}=\frac{\sqrt{1319}i}{4}$$ $$n+\frac{23}{4}=-\frac{\sqrt{1319}i}{4}$$

Subtract $\frac{23}{4}$ from both sides of the equation.

$$n=\frac{-23+\sqrt{1319}i}{4}$$ $$n=\frac{-\sqrt{1319}i-23}{4}$$