Solve the Quadratic Equation: Balancing and Simplifying to Find x

Solve the following equation:

$-2x^2+6x-12=-4x^2+19x-5$

To solve this quadratic equation, let us first simplify and rearrange the terms:

Start with the original equation:
$-2x^2 + 6x - 12 = -4x^2 + 19x - 5$

Move all terms to one side to form a standard quadratic equation by adding $4x^2$, subtracting $19x$, and adding $5$ to both sides:

$(-2x^2 + 6x - 12) + 4x^2 - 19x + 5 = 0$

This simplifies to:
$2x^2 - 13x - 7 = 0$

Now, identify the coefficients $a = 2$, $b = -13$, and $c = -7$.

Apply the quadratic formula:
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Subsitute $a$, $b$, and $c$ into the formula:

$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2}$

Simplify:
$x = \frac{13 \pm \sqrt{169 + 56}}{4}$

$x = \frac{13 \pm \sqrt{225}}{4}$

The square root of 225 is 15, thus:

$x = \frac{13 \pm 15}{4}$

Calculate the two possible solutions:

• First solution: $x_1 = \frac{13 + 15}{4} = \frac{28}{4} = 7$
• Second solution: $x_2 = \frac{13 - 15}{4} = \frac{-2}{4} = -\frac{1}{2}$

Therefore, the solutions to the problem are $x_1 = 7$ and $x_2 = -\frac{1}{2}$.

Thus, the correct answer is option 2: $x_1=7$, $x_2=-\frac{1}{2}$.