You can start by finding the solutions to \(x^2+5x+3 \). When you plug that into the quadratic formula, you will end up with \(\frac{-5 \pm \sqrt{13}}{2}\).\(\left(\frac{-5+\sqrt{13}}{2}\right)^2=38-10\sqrt{13}\) and \(\left(\frac{-5-\sqrt{13}}{2}\right)^2=38+10\sqrt{13}\) since those are the solutions to the second quadratic, we know that the second quadratic is the expanded form of \((x-38+10\sqrt{13})(x-38-10\sqrt{13})\). When you expand that, you get \(x^2-76x+144 \). \(b=-76\), and \(c=144 \), so \(b+c=\color{red}{68}\).
