Note that all points on the parabola \(y = x^2 + x\) are of the form \((x, x^2 + x)\)

Assume that the tangent at \((a, a^2 + a)\) passes through (2, -3)

We know that the slope of tangent at any point is the derivative at that point.

Differentiate \(y = x^2 + x, To\ get \frac{dy}{dx} = 2x + 1\)

Therefore, the slope of tangent at \((a, a^2 + a) is 2a + 1 \rightarrow (1)\)

Since the tangent passes through the points \((a, a^2 + a)\) and (2. -3). We can write the slope of the tangent as

\(\frac{(a^2 + a)-(-3)}{a - 2} = \frac{a^2 + a + 3}{a - 2} \rightarrow (2)\)

Using (1) and (2), we can write \(\frac{a^2 + a + 3}{a - 2} = 2a + 1\)

\(a^2 + a + 3 = (2a + 1)(a - 2)\)

\(a^2 + a + 3 = 2a^2 + a - 4a - 2\)

\(a^2 - 4a - 5 = 0\)

\((a + 1)(a - 5) = 0 \Rightarrow a = -1\ and\ a = 5\)

Therefore, there are two tangents that pass through (2, -3)

Finding the equation of the tangent corresponding to \(a = -1\)

Note that the slope of the tangent is \(2a + 1 = 2(-1) + 1 = -1\)

Since the tangent. passes through (2, -3).

Therefore, the equation of the tangent is

\(\frac{y - (-3)}{x - 2} = -1\)

\(y + 3 = -(x - 2)\)

\(x + y + 1 = 0\)

Finding the equation of the tangent corresponding to \(a = 5\)

Note that the slope of the tangent is \(2a + 1 = 2(5) + 1 = 11\)

Since the tangent. passes through (2, -3).

Therefore, the equation of the tangent is

\(\frac{y - (-3)}{x - 2} = 11\)

\(y + 3 = 11(x - 2)\) \(y = 11x - 25\)