The line ${\ell }_1$ passes through the point $(-4,0)$ and has gradient $m.$

${\ell }_1$ intersects the curve

in exactly three points.

Given that $m$ is of the form $a-b\sqrt{c}$ where $a,b,c\in \mathbb{N},$ find the exact value of $m.$

Try moving $m$ around in the picture above.

It suggests there is a value of $m$ between $0.5$ and $0.6$ where we have exactly three points of intersection.

Because we know $(x_0,y_0)=(-4,0)$ lies on the line, the equation of the line can be expressed, using the point-slope formula, as

Simplify this before using it.

The key is to make sure $y=2-x^2$ (that's the $n$-shaped part of the parabola) and the line $y=mx+4m$ intersect exactly once.

Use the discriminant.