\(x^2-4x-5=0\)

\begin{align}\\ (x-5)(x+1) & =0\\ x= 5\quad or \quad x=-1\\ \end{align}

\(4x^2-3x-10=0\)

\begin{align}4x^2-8x+5x-10 & =0\\ 4x(x-2)+5(x-2) & =0\\ (4x+5)(x-2) & = 0\\ x=-\frac{5}{4}\quad or \quad x=2\\ \end{align}

\(36x^2-9=0\)

\begin{align}\\ (6x+3)(6x-3) & =0\\ x=-\frac{1}{2} \quad or \quad x=\frac{1}{2}\\ \end{align}

\(2x^2+6x+3=0\) (Answer to 2d.p.)

\begin{gather}Using\quad the\quad quadratic\quad formula\\ \frac{-b\;{\displaystyle\pm}\sqrt{b^2-4ac}}{2a}\\\\ x =\frac{-6\;{\displaystyle\pm}\sqrt{6^2-4(2)(3)}}{2(2)}\\ x =\frac{-6\;{\displaystyle\pm}\sqrt{12}}{4}\\ x=-0.63 \quad or \quad x= -2.37\\ \end{gather}