Solve (x+4)(3x-1/4) = 3(x²+5): Complete Quadratic Equation Guide

To solve the equation $(x+4)(3x-\frac{1}{4}) = 3(x^2+5)$, follow these steps:

• Step 1: Expand the left side of the equation
  $(x + 4)(3x - \frac{1}{4})$

Using the distributive property:

$x(3x) + x(-\frac{1}{4}) + 4(3x) + 4(-\frac{1}{4})$

$= 3x^2 - \frac{x}{4} + 12x - 1$

• Step 2: Simplify the expanded left side
  Combine like terms:

$3x^2 + \left(12x - \frac{x}{4}\right) - 1$

Convert $\frac{x}{4}$ to a common denominator: $\frac{48x}{4} - \frac{x}{4} = \frac{47x}{4}$

Thus, the left side is: $3x^2 + \frac{47x}{4} - 1$

• Step 3: Simplify the right side
  $3(x^2 + 5)$

$= 3x^2 + 15$

• Step 4: Set the simplified expressions equal and solve for $x$

$3x^2 + \frac{47x}{4} - 1 = 3x^2 + 15$

Subtract $3x^2$ from both sides:

$\frac{47x}{4} - 1 = 15$

Add 1 to both sides:

$\frac{47x}{4} = 16$

Multiply both sides by 4 to clear the fraction:

$47x = 64$

• Step 5: Solve for $x$

$x = \frac{64}{47}$

Express $\frac{64}{47}$ as a mixed number:

$x = 1\frac{17}{47}$

Therefore, the solution to the equation is $x = 1\frac{17}{47}$.