Solve 15(x)^(2)=-7x+1-3x-14 | 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

$$15 { x }^{ 2 } =-7x+1-3x-14$$

Solve for x (complex solution)

$x=\frac{\sqrt{170}i}{15}-\frac{1}{3}\approx -0.333333333+0.869226987i$
$x=-\frac{\sqrt{170}i}{15}-\frac{1}{3}\approx -0.333333333-0.869226987i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Combine $-7x$ and $-3x$ to get $-10x$.

$$15x^{2}=-10x+1-14$$

Subtract $14$ from $1$ to get $-13$.

$$15x^{2}=-10x-13$$

Add $10x$ to both sides.

$$15x^{2}+10x=-13$$

Add $13$ to both sides.

$$15x^{2}+10x+13=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $15$ for $a$, $10$ for $b$, and $13$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-10±\sqrt{10^{2}-4\times 15\times 13}}{2\times 15}$$

Square $10$.

$$x=\frac{-10±\sqrt{100-4\times 15\times 13}}{2\times 15}$$

Multiply $-4$ times $15$.

$$x=\frac{-10±\sqrt{100-60\times 13}}{2\times 15}$$

Multiply $-60$ times $13$.

$$x=\frac{-10±\sqrt{100-780}}{2\times 15}$$

Add $100$ to $-780$.

$$x=\frac{-10±\sqrt{-680}}{2\times 15}$$

Take the square root of $-680$.

$$x=\frac{-10±2\sqrt{170}i}{2\times 15}$$

Multiply $2$ times $15$.

$$x=\frac{-10±2\sqrt{170}i}{30}$$

Now solve the equation $x=\frac{-10±2\sqrt{170}i}{30}$ when $±$ is plus. Add $-10$ to $2i\sqrt{170}$.

$$x=\frac{-10+2\sqrt{170}i}{30}$$

Divide $-10+2i\sqrt{170}$ by $30$.

$$x=\frac{\sqrt{170}i}{15}-\frac{1}{3}$$

Now solve the equation $x=\frac{-10±2\sqrt{170}i}{30}$ when $±$ is minus. Subtract $2i\sqrt{170}$ from $-10$.

$$x=\frac{-2\sqrt{170}i-10}{30}$$

Divide $-10-2i\sqrt{170}$ by $30$.

$$x=-\frac{\sqrt{170}i}{15}-\frac{1}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{170}i}{15}-\frac{1}{3}$$ $$x=-\frac{\sqrt{170}i}{15}-\frac{1}{3}$$

Steps for Completing the Square

Combine $-7x$ and $-3x$ to get $-10x$.

$$15x^{2}=-10x+1-14$$

Subtract $14$ from $1$ to get $-13$.

$$15x^{2}=-10x-13$$

Add $10x$ to both sides.

$$15x^{2}+10x=-13$$

Divide both sides by $15$.

$$\frac{15x^{2}+10x}{15}=-\frac{13}{15}$$

Dividing by $15$ undoes the multiplication by $15$.

$$x^{2}+\frac{10}{15}x=-\frac{13}{15}$$

Reduce the fraction $\frac{10}{15}$ to lowest terms by extracting and canceling out $5$.

$$x^{2}+\frac{2}{3}x=-\frac{13}{15}$$

Divide $\frac{2}{3}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{3}$. Then add the square of $\frac{1}{3}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{13}{15}+\left(\frac{1}{3}\right)^{2}$$

Square $\frac{1}{3}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{13}{15}+\frac{1}{9}$$

Add $-\frac{13}{15}$ to $\frac{1}{9}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{34}{45}$$

Factor $x^{2}+\frac{2}{3}x+\frac{1}{9}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{34}{45}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{34}{45}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{170}i}{15}$$ $$x+\frac{1}{3}=-\frac{\sqrt{170}i}{15}$$

Subtract $\frac{1}{3}$ from both sides of the equation.

$$x=\frac{\sqrt{170}i}{15}-\frac{1}{3}$$ $$x=-\frac{\sqrt{170}i}{15}-\frac{1}{3}$$