Solve $-10\frac{d^{2}y\left(x\right)}{d{x}^2}=$ $7y\left(x\right)$.

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$.
Substitute $y\left(x\right)=$ $e^{\lambda x}$ into the differential equation:
$-10\frac{d^2}{d{x}^2}\left(e^{\lambda x}\right)-$ $7e^{\lambda x}=$ $0$

Substitute $\frac{d^2}{d{x}^2}\left(e^{\lambda x}\right)=$ ${\lambda }^2{e}^{\lambda x}$:
$-10{\lambda }^2{e}^{\lambda x}-$ $7e^{\lambda x}=$ $0$

Factor out $e^{\lambda x}$:
$\left(-10{\lambda }^2-7\right)e^{\lambda x}=$ $0$

Since $e^{\lambda x}\ne 0$ for any finite $\lambda$, the zeros must come from the polynomial:
$-10{\lambda }^2-$ $7=$ $0$

Solve for $\lambda$:
$\lambda =$ $i\sqrt{\frac{7}{10}}$ or $\lambda =$ $-i\sqrt{\frac{7}{10}}$

The roots $\lambda =$ $\pm i\sqrt{\frac{7}{10}}$ give $y_1\left(x\right)=$ $c_1{e}^{i\sqrt{\frac{7}{10}}x}$ and $y_2\left(x\right)=$ $c_2{e}^{-i\sqrt{\frac{7}{10}}x}$ as solutions, where $c_1$ and $c_2$ are arbitrary constants.
The general solution is the sum of the above solutions:
$y\left(x\right)=$ $y_1\left(x\right)+$ $y_2\left(x\right)=$ $c_1{e}^{i\sqrt{\frac{7}{10}}x}+$ $c_2{e}^{-i\sqrt{\frac{7}{10}}x}$

Apply Euler's identity $e^{\alpha +i\beta }=$ $e^{\alpha }\cos \left(\beta \right)+$ $i{e}^{\alpha }\sin \left(\beta \right)$:
$y\left(x\right)=$ $c_1\left(\cos \left(\sqrt{\frac{7}{10}}x\right)+i\sin \left(\sqrt{\frac{7}{10}}x\right)\right)+$ $c_2\left(\cos \left(\sqrt{\frac{7}{10}}x\right)-i\sin \left(\sqrt{\frac{7}{10}}x\right)\right)$

Regroup terms:
$y\left(x\right)=$ $\left(c_1+c_2\right)\cos \left(\sqrt{\frac{7}{10}}x\right)+$ $i\left(c_1-c_2\right)\sin \left(\sqrt{\frac{7}{10}}x\right)$

Redefine $c_1+c_2$ as $k_1$ and $i\left(c_1-c_2\right)$ as $k_2$ since these are arbitrary constants, so:
$y\left(x\right)=$ $k_1\cos \left(\sqrt{\frac{7}{10}}x\right)+$ $k_2\sin \left(\sqrt{\frac{7}{10}}x\right)$