(x+7)^4+(x+7)^2-30=0 equation

Given the equation:
$$\left(x + 7\right)^{4} + \left(x + 7\right)^{2} - 30 = 0$$
Do replacement
$$v = \left(x + 7\right)^{2}$$
then the equation will be the:
$$v^{2} + v - 30 = 0$$
This equation is of the form
$$a*v^2 + b*v + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = 1$$
$$c = -30$$
, then
$$D = b^2 - 4 * a * c = $$
$$1^{2} - 1 \cdot 4 \left(-30\right) = 121$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 5$$
Simplify
$$v_{2} = -6$$
Simplify
The final answer:
Because
$$v = \left(x + 7\right)^{2}$$
then
$$x_{1} = \sqrt{v_{1}} - 7$$
$$x_{2} = - \sqrt{v_{1}} - 7$$
$$x_{3} = \sqrt{v_{2}} - 7$$
$$x_{4} = - \sqrt{v_{2}} - 7$$
then:
$$x_{1} = - \frac{7}{1} + \frac{1 \cdot 5^{\frac{1}{2}}}{1} = -7 + \sqrt{5}$$
$$x_{2} = - \frac{7}{1} + \frac{\left(-1\right) 5^{\frac{1}{2}}}{1} = -7 - \sqrt{5}$$
$$x_{3} = - \frac{7}{1} + \frac{1 \left(-6\right)^{\frac{1}{2}}}{1} = -7 + \sqrt{6} i$$
$$x_{4} = - \frac{7}{1} + \frac{\left(-1\right) \left(-6\right)^{\frac{1}{2}}}{1} = -7 - \sqrt{6} i$$