Solve the Quadratic Equation: Find x in 4x² + 9x - 5 = 7 - 4x

Solve the following equation:

$4x^2+9x-5=7-4x$

Let's solve the equation $4x^2 + 9x - 5 = 7 - 4x$ step by step:

• Step 1: Move all terms to one side to form a standard quadratic equation.

We begin by subtracting $7$ and adding $4x$ to both sides of the given equation:

$4x^2 + 9x - 5 - 7 + 4x = 0$

Combining like terms, we get:

$4x^2 + 13x - 12 = 0$

• Step 2: Identify the coefficients $a$, $b$, and $c$ in the quadratic equation $4x^2 + 13x - 12 = 0$.

Here, $a = 4$, $b = 13$, and $c = -12$.

• Step 3: Apply the quadratic formula:

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Plugging in the values of $a$, $b$, and $c$:

$x = \frac{-13 \pm \sqrt{13^2 - 4 \times 4 \times (-12)}}{2 \times 4}$

$x = \frac{-13 \pm \sqrt{169 + 192}}{8}$

$x = \frac{-13 \pm \sqrt{361}}{8}$

$x = \frac{-13 \pm 19}{8}$

• Step 4: Solve for the possible values of $x$.

We have two solutions:

For the positive case:

$x_1 = \frac{-13 + 19}{8} = \frac{6}{8} = \frac{3}{4}$

For the negative case:

$x_2 = \frac{-13 - 19}{8} = \frac{-32}{8} = -4$

Therefore, the solutions are $x_1 = \frac{3}{4}$ and $x_2 = -4$.