Solve 6n ^ 2 + 3 = - 10n | 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

$$6n^{2}+3=-10n$$

Solve for n

$n=\frac{\sqrt{7}-5}{6}\approx -0.392374781$
$n=\frac{-\sqrt{7}-5}{6}\approx -1.274291885$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Add $10n$ to both sides.

$$6n^{2}+3+10n=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$6n^{2}+10n+3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $6$ for $a$, $10$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$n=\frac{-10±\sqrt{10^{2}-4\times 6\times 3}}{2\times 6}$$

Square $10$.

$$n=\frac{-10±\sqrt{100-4\times 6\times 3}}{2\times 6}$$

Multiply $-4$ times $6$.

$$n=\frac{-10±\sqrt{100-24\times 3}}{2\times 6}$$

Multiply $-24$ times $3$.

$$n=\frac{-10±\sqrt{100-72}}{2\times 6}$$

Add $100$ to $-72$.

$$n=\frac{-10±\sqrt{28}}{2\times 6}$$

Take the square root of $28$.

$$n=\frac{-10±2\sqrt{7}}{2\times 6}$$

Multiply $2$ times $6$.

$$n=\frac{-10±2\sqrt{7}}{12}$$

Now solve the equation $n=\frac{-10±2\sqrt{7}}{12}$ when $±$ is plus. Add $-10$ to $2\sqrt{7}$.

$$n=\frac{2\sqrt{7}-10}{12}$$

Divide $-10+2\sqrt{7}$ by $12$.

$$n=\frac{\sqrt{7}-5}{6}$$

Now solve the equation $n=\frac{-10±2\sqrt{7}}{12}$ when $±$ is minus. Subtract $2\sqrt{7}$ from $-10$.

$$n=\frac{-2\sqrt{7}-10}{12}$$

Divide $-10-2\sqrt{7}$ by $12$.

$$n=\frac{-\sqrt{7}-5}{6}$$

The equation is now solved.

$$n=\frac{\sqrt{7}-5}{6}$$ $$n=\frac{-\sqrt{7}-5}{6}$$

Steps for Completing the Square

Add $10n$ to both sides.

$$6n^{2}+3+10n=0$$

Subtract $3$ from both sides. Anything subtracted from zero gives its negation.

$$6n^{2}+10n=-3$$

Divide both sides by $6$.

$$\frac{6n^{2}+10n}{6}=-\frac{3}{6}$$

Dividing by $6$ undoes the multiplication by $6$.

$$n^{2}+\frac{10}{6}n=-\frac{3}{6}$$

Reduce the fraction $\frac{10}{6}$ to lowest terms by extracting and canceling out $2$.

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

Reduce the fraction $\frac{-3}{6}$ to lowest terms by extracting and canceling out $3$.

$$n^{2}+\frac{5}{3}n=-\frac{1}{2}$$

Divide $\frac{5}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{5}{6}$. Then add the square of $\frac{5}{6}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{5}{3}n+\left(\frac{5}{6}\right)^{2}=-\frac{1}{2}+\left(\frac{5}{6}\right)^{2}$$

Square $\frac{5}{6}$ by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{5}{3}n+\frac{25}{36}=-\frac{1}{2}+\frac{25}{36}$$

Add $-\frac{1}{2}$ to $\frac{25}{36}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{5}{3}n+\frac{25}{36}=\frac{7}{36}$$

Factor $n^{2}+\frac{5}{3}n+\frac{25}{36}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(n+\frac{5}{6}\right)^{2}=\frac{7}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{5}{6}\right)^{2}}=\sqrt{\frac{7}{36}}$$

Simplify.

$$n+\frac{5}{6}=\frac{\sqrt{7}}{6}$$ $$n+\frac{5}{6}=-\frac{\sqrt{7}}{6}$$

Subtract $\frac{5}{6}$ from both sides of the equation.

$$n=\frac{\sqrt{7}-5}{6}$$ $$n=\frac{-\sqrt{7}-5}{6}$$