Solve ер १ 4 + - = 21 | 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

$$\frac { n } { 2 } \frac { - 3 n } { 4 } + \frac { 5 n } { 6 } = 91$$

Solve for n

$n=\frac{10+2\sqrt{4889}i}{9}\approx 1.111111111+15.538085428i$
$n=\frac{-2\sqrt{4889}i+10}{9}\approx 1.111111111-15.538085428i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $12$, the least common multiple of $2,4,6$.

$$6n\times \frac{-3n}{4}+2\times 5n=1092$$

Express $6\times \frac{-3n}{4}$ as a single fraction.

$$\frac{6\left(-3\right)n}{4}n+2\times 5n=1092$$

Multiply $2$ and $5$ to get $10$.

$$\frac{6\left(-3\right)n}{4}n+10n=1092$$

Multiply $6$ and $-3$ to get $-18$.

$$\frac{-18n}{4}n+10n=1092$$

Divide $-18n$ by $4$ to get $-\frac{9}{2}n$.

$$-\frac{9}{2}nn+10n=1092$$

Multiply $n$ and $n$ to get $n^{2}$.

$$-\frac{9}{2}n^{2}+10n=1092$$

Subtract $1092$ from both sides.

$$-\frac{9}{2}n^{2}+10n-1092=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-\frac{9}{2}$ for $a$, $10$ for $b$, and $-1092$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{-10±\sqrt{10^{2}-4\left(-\frac{9}{2}\right)\left(-1092\right)}}{2\left(-\frac{9}{2}\right)}$$

Square $10$.

$$n=\frac{-10±\sqrt{100-4\left(-\frac{9}{2}\right)\left(-1092\right)}}{2\left(-\frac{9}{2}\right)}$$

Multiply $-4$ times $-\frac{9}{2}$.

$$n=\frac{-10±\sqrt{100+18\left(-1092\right)}}{2\left(-\frac{9}{2}\right)}$$

Multiply $18$ times $-1092$.

$$n=\frac{-10±\sqrt{100-19656}}{2\left(-\frac{9}{2}\right)}$$

Add $100$ to $-19656$.

$$n=\frac{-10±\sqrt{-19556}}{2\left(-\frac{9}{2}\right)}$$

Take the square root of $-19556$.

$$n=\frac{-10±2\sqrt{4889}i}{2\left(-\frac{9}{2}\right)}$$

Multiply $2$ times $-\frac{9}{2}$.

$$n=\frac{-10±2\sqrt{4889}i}{-9}$$

Now solve the equation $n=\frac{-10±2\sqrt{4889}i}{-9}$ when $±$ is plus. Add $-10$ to $2i\sqrt{4889}$.

$$n=\frac{-10+2\sqrt{4889}i}{-9}$$

Divide $-10+2i\sqrt{4889}$ by $-9$.

$$n=\frac{-2\sqrt{4889}i+10}{9}$$

Now solve the equation $n=\frac{-10±2\sqrt{4889}i}{-9}$ when $±$ is minus. Subtract $2i\sqrt{4889}$ from $-10$.

$$n=\frac{-2\sqrt{4889}i-10}{-9}$$

Divide $-10-2i\sqrt{4889}$ by $-9$.

$$n=\frac{10+2\sqrt{4889}i}{9}$$

The equation is now solved.

$$n=\frac{-2\sqrt{4889}i+10}{9}$$ $$n=\frac{10+2\sqrt{4889}i}{9}$$

Steps for Completing the Square

Multiply both sides of the equation by $12$, the least common multiple of $2,4,6$.

$$6n\times \frac{-3n}{4}+2\times 5n=1092$$

Express $6\times \frac{-3n}{4}$ as a single fraction.

$$\frac{6\left(-3\right)n}{4}n+2\times 5n=1092$$

Multiply $2$ and $5$ to get $10$.

$$\frac{6\left(-3\right)n}{4}n+10n=1092$$

Multiply $6$ and $-3$ to get $-18$.

$$\frac{-18n}{4}n+10n=1092$$

Divide $-18n$ by $4$ to get $-\frac{9}{2}n$.

$$-\frac{9}{2}nn+10n=1092$$

Multiply $n$ and $n$ to get $n^{2}$.

$$-\frac{9}{2}n^{2}+10n=1092$$

Divide both sides of the equation by $-\frac{9}{2}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{9}{2}n^{2}+10n}{-\frac{9}{2}}=\frac{1092}{-\frac{9}{2}}$$

Dividing by $-\frac{9}{2}$ undoes the multiplication by $-\frac{9}{2}$.

$$n^{2}+\frac{10}{-\frac{9}{2}}n=\frac{1092}{-\frac{9}{2}}$$

Divide $10$ by $-\frac{9}{2}$ by multiplying $10$ by the reciprocal of $-\frac{9}{2}$.

$$n^{2}-\frac{20}{9}n=\frac{1092}{-\frac{9}{2}}$$

Divide $1092$ by $-\frac{9}{2}$ by multiplying $1092$ by the reciprocal of $-\frac{9}{2}$.

$$n^{2}-\frac{20}{9}n=-\frac{728}{3}$$

Divide $-\frac{20}{9}$, the coefficient of the $x$ term, by $2$ to get $-\frac{10}{9}$. Then add the square of $-\frac{10}{9}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{20}{9}n+\left(-\frac{10}{9}\right)^{2}=-\frac{728}{3}+\left(-\frac{10}{9}\right)^{2}$$

Square $-\frac{10}{9}$ by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{20}{9}n+\frac{100}{81}=-\frac{728}{3}+\frac{100}{81}$$

Add $-\frac{728}{3}$ to $\frac{100}{81}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{20}{9}n+\frac{100}{81}=-\frac{19556}{81}$$

Factor $n^{2}-\frac{20}{9}n+\frac{100}{81}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(n-\frac{10}{9}\right)^{2}=-\frac{19556}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{10}{9}\right)^{2}}=\sqrt{-\frac{19556}{81}}$$

Simplify.

$$n-\frac{10}{9}=\frac{2\sqrt{4889}i}{9}$$ $$n-\frac{10}{9}=-\frac{2\sqrt{4889}i}{9}$$

Add $\frac{10}{9}$ to both sides of the equation.

$$n=\frac{10+2\sqrt{4889}i}{9}$$ $$n=\frac{-2\sqrt{4889}i+10}{9}$$