Ordinary least squares estimators

Write $Q(b_0,b_1)=\sum_{i=1}^n (y_i-b_0-b_1x_i)^2$. Taking the partial derivative with respect to $b_0$ and setting it to zero: $\dfrac{\partial Q}{\partial b_0}=-2\sum_{i=1}^n(y_i-b_0-b_1x_i)=0$, which simplifies to the first normal equation $\sum_{i=1}^n y_i = nb_0+b_1\sum_{i=1}^n x_i$, i.e. $b_0=\bar y-b_1\bar x$.

Taking the partial derivative with respect to $b_1$ and setting it to zero: $\dfrac{\partial Q}{\partial b_1}=-2\sum_{i=1}^n x_i(y_i-b_0-b_1x_i)=0$, the second normal equation $\sum_{i=1}^n x_iy_i = b_0\sum_{i=1}^n x_i+b_1\sum_{i=1}^n x_i^2$.

Substitute $b_0=\bar y-b_1\bar x$ from the first equation into the second: $\sum_{i=1}^n x_iy_i = (\bar y-b_1\bar x)n\bar x+b_1\sum_{i=1}^n x_i^2 = n\bar x\bar y+b_1\left(\sum_{i=1}^n x_i^2-n\bar x^2\right)$. Rearranging isolates $b_1$: $b_1\left(\sum_{i=1}^n x_i^2-n\bar x^2\right)=\sum_{i=1}^n x_iy_i-n\bar x\bar y$.

A direct algebraic expansion shows $\sum_{i=1}^n x_i^2-n\bar x^2=\sum_{i=1}^n(x_i-\bar x)^2=S_{xx}$ and $\sum_{i=1}^n x_iy_i-n\bar x\bar y=\sum_{i=1}^n(x_i-\bar x)(y_i-\bar y)=S_{xy}$, so $b_1=S_{xy}/S_{xx}$. Substituting back gives $b_0=\bar y-b_1\bar x$. Because $Q$ is a sum of squares in a quadratic form that grows without bound as $|b_0|,|b_1|\to\infty$, this unique stationary point must be the global minimum.