x+2x²-4=8+3x²-7x equation

Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$2 x^{2} + x - 4 = 3 x^{2} - 7 x + 8$$
to
$$\left(- 3 x^{2} + 7 x - 8\right) + \left(2 x^{2} + x - 4\right) = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = -1$$
$$b = 8$$
$$c = -12$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) \left(\left(-1\right) 4\right) \left(-12\right) + 8^{2} = 16$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 2$$
Simplify
$$x_{2} = 6$$
Simplify