solving a quadratic in surd form I am trying to come up with a surd solution for this quadratic: $$-2s^2-s+3=0$$ I can only seem to find a roots in decimal form, namely $x = \frac{-2}{3}$ or $x = 1$ The question specifies surd form but i can not do it. any help showing the methodology would be highly appreciated.

$f(s) = -2s^2-s+3 =0\implies s^2+\frac{s}{2}-\frac{3}{2}=0$

Then $$(s+\frac{1}{4})^2-\frac{3}{2}-\frac{1}{16}=0$$

$$(s+\frac{1}{4})^2=\frac{1+24}{16}=\frac{25}{16}$$

$$s=\frac{-1\pm 5}{4}\implies s= -\frac{3}{2} \quad\text{or}\quad s=1$$

If you insist on writing them inside a square root, then:

$s= -\sqrt{\frac{9}{4}}\quad\text{or}\quad s=\sqrt{1}$

Note in general that a quadratic of the form $f(x)=ax^2+bx+c$, can be written as:

$f(x)=a[x^2+\frac{b}{a}x+\frac{c}{a}]=a[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}]$

This is known as completing the square, it is useful because setting $f(x)=0$:

$(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a} =\frac{b^2-4ac}{4a^2}$

$x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\quad$ which you should recognise as the quadratic formula. Completing the square is an alternative to just substituting the coefficients into this formula.