Let \(\lambda = -4\) be an eigenvalue of the matrix \[A=\matrix{-25 & 28 \\ -21 & 24}\] is. Then there must be a vector \(\vec{v}\) such that \(A\vec{v}=-4\vec{v}\), that is, \[(A+4 I)\vec{v}=\vec{0}\text.\] In other words, we must find the kernel of the matrix \(A+4 I\).
We can do this through row reduction of the matrix \[A+4 I = \matrix{-25 & 28 \\ -21 & 24} - \matrix{-4 & 0 \\0& -4 }=\matrix{ -21 & 28 \\ -21 & 28}\] This can be done as follows:
\[\begin{aligned}
\matrix{-21&28\\-21&28\\}&\sim\matrix{1&-{{4}\over{3}}\\-21&28\\}&{\blue{\begin{array}{c}-{{1}\over{21}}R_1\\\phantom{x}\end{array}}}\\\\ &\sim\matrix{1&-{{4}\over{3}}\\0&0\\}&{\blue{\begin{array}{c}\phantom{x}\\R_2+21R_1\end{array}}} \end{aligned}\] So the eigenspace for \(\lambda = -4\) equals \(\left\{ r \cv{-4\\-3} \middle|\;r\in\mathbb R\right\}\).
If needed, we avoided here fractions in the solution.

In other words, the eigenspace of eigenvalue \(-4\) equals \(\left\langle\cv{-4\\-3}\right\rangle\)