[tex]\text{The vertex of}\ ax^2+bx+c:\\\\(h,\ k),\ \text{where}\ h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(-b^2-4ac)}{4a}[/tex]
[tex]\text{We have:}\\f(x)=x^2+5x-3\\\\a=1,\ b=5,\ c=-3\\\\h=\dfrac{-5}{2\cdot1}=\dfrac{-5}{2}=-2.5\\\\k=f(-2.5)=(-2.5)^2+5(-2.5)-3=6.25-12.5-3=-9.25[/tex]
[tex]\text{Answer: the vertex is in}\ (-2.5;\ -9.25)[/tex]
