Make a quadratic equation with one x intercept, -4?

If you want only one x-intercept at point P( -4| 0 ) with the x-axis as tangent of the parabola, derived from the vortex-form a(x - e) + f, where e is the x-coordinate and f is the y-coordinate of the vortex. As we know the vortex, we can simplify that :

f(x) = a(x + 4)^2
f(x) = a( x^2 + 8x + 16)
f(-4) = a( 16 - 32 + 16) = 0

If instead you want two points, where the parabola cuts through the x-axis, one of which is at P_1 ( -4 | 0), and the other at P_2 ( x_2 | 0 ) :

y = f(x) = a(x + b)^2 + c

Work out f(-4) with an arbitrary b and adjust c to make f(-4) = 0. Finally you can multiply an arbitrary a into the parabola and if you want to know the x_2-coordinate, solve the corresponding quadratic equation by e. g. factorising.

Example: b = 6

f(x) = ( x + 6 )^2 + c
f(-4) = ( -4 + 6 )^2 + c = 0
f(-4) = 4 + c = 0
Adjust c : c= -4
f(x) = x^2 + 12x + 36 - 4
f(x) = x^2 + 12x + 32
f(-4) = 16 - 48 + 32 = 0

Factorise :
f(x) = (x + 4)(x + 8)
x_1 = -4, x_2 = -8