Solve 13 : - 14 : - 3x + + 2 x - 3 2x + 4 | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

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

a⋅(b-2)=3b

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

a⋅(b-2)=3b

Question

$$14 : - \quad \frac { x - 3 } { 2 } = \frac { 2 x + 4 } { 5 }$$

Solve for x (complex solution)

$x=\frac{-\sqrt{255}i+1}{2}\approx 0.5-7.984359711i$
$x=\frac{1+\sqrt{255}i}{2}\approx 0.5+7.984359711i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $5$.

$$-10\left(x-3\right)^{-1}\times 14=2x+4$$

Multiply $-10$ and $14$ to get $-140$.

$$-140\left(x-3\right)^{-1}=2x+4$$

Subtract $2x$ from both sides.

$$-140\left(x-3\right)^{-1}-2x=4$$

Subtract $4$ from both sides.

$$-140\left(x-3\right)^{-1}-2x-4=0$$

Reorder the terms.

$$-2x-4-140\times \frac{1}{x-3}=0$$

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$-2x\left(x-3\right)+\left(x-3\right)\left(-4\right)-140=0$$

Use the distributive property to multiply $-2x$ by $x-3$.

$$-2x^{2}+6x+\left(x-3\right)\left(-4\right)-140=0$$

Use the distributive property to multiply $x-3$ by $-4$.

$$-2x^{2}+6x-4x+12-140=0$$

Combine $6x$ and $-4x$ to get $2x$.

$$-2x^{2}+2x+12-140=0$$

Subtract $140$ from $12$ to get $-128$.

$$-2x^{2}+2x-128=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-2$ for $a$, $2$ for $b$, and $-128$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-2\right)\left(-128\right)}}{2\left(-2\right)}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\left(-2\right)\left(-128\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-2±\sqrt{4+8\left(-128\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-128$.

$$x=\frac{-2±\sqrt{4-1024}}{2\left(-2\right)}$$

Add $4$ to $-1024$.

$$x=\frac{-2±\sqrt{-1020}}{2\left(-2\right)}$$

Take the square root of $-1020$.

$$x=\frac{-2±2\sqrt{255}i}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-2±2\sqrt{255}i}{-4}$$

Now solve the equation $x=\frac{-2±2\sqrt{255}i}{-4}$ when $±$ is plus. Add $-2$ to $2i\sqrt{255}$.

$$x=\frac{-2+2\sqrt{255}i}{-4}$$

Divide $-2+2i\sqrt{255}$ by $-4$.

$$x=\frac{-\sqrt{255}i+1}{2}$$

Now solve the equation $x=\frac{-2±2\sqrt{255}i}{-4}$ when $±$ is minus. Subtract $2i\sqrt{255}$ from $-2$.

$$x=\frac{-2\sqrt{255}i-2}{-4}$$

Divide $-2-2i\sqrt{255}$ by $-4$.

$$x=\frac{1+\sqrt{255}i}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{255}i+1}{2}$$ $$x=\frac{1+\sqrt{255}i}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by $5$.

$$-10\left(x-3\right)^{-1}\times 14=2x+4$$

Multiply $-10$ and $14$ to get $-140$.

$$-140\left(x-3\right)^{-1}=2x+4$$

Subtract $2x$ from both sides.

$$-140\left(x-3\right)^{-1}-2x=4$$

Reorder the terms.

$$-2x-140\times \frac{1}{x-3}=4$$

Variable $x$ cannot be equal to $3$ since division by zero is not defined. Multiply both sides of the equation by $x-3$.

$$-2x\left(x-3\right)-140=4\left(x-3\right)$$

Use the distributive property to multiply $-2x$ by $x-3$.

$$-2x^{2}+6x-140=4\left(x-3\right)$$

Use the distributive property to multiply $4$ by $x-3$.

$$-2x^{2}+6x-140=4x-12$$

Subtract $4x$ from both sides.

$$-2x^{2}+6x-140-4x=-12$$

Combine $6x$ and $-4x$ to get $2x$.

$$-2x^{2}+2x-140=-12$$

Add $140$ to both sides.

$$-2x^{2}+2x=-12+140$$

Add $-12$ and $140$ to get $128$.

$$-2x^{2}+2x=128$$

Divide both sides by $-2$.

$$\frac{-2x^{2}+2x}{-2}=\frac{128}{-2}$$

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

$$x^{2}+\frac{2}{-2}x=\frac{128}{-2}$$

Divide $2$ by $-2$.

$$x^{2}-x=\frac{128}{-2}$$

Divide $128$ by $-2$.

$$x^{2}-x=-64$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-64+\left(-\frac{1}{2}\right)^{2}$$

Square $-\frac{1}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-64+\frac{1}{4}$$

Add $-64$ to $\frac{1}{4}$.

$$x^{2}-x+\frac{1}{4}=-\frac{255}{4}$$

Factor $x^{2}-x+\frac{1}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{255}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{255}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{255}i}{2}$$ $$x-\frac{1}{2}=-\frac{\sqrt{255}i}{2}$$

Add $\frac{1}{2}$ to both sides of the equation.

$$x=\frac{1+\sqrt{255}i}{2}$$ $$x=\frac{-\sqrt{255}i+1}{2}$$