$f\left(x\right)=x^2+8x-20={(x+4)}^2-36$, vertex $(-4,-36)$.

${(x+4)}^2-36=0$ gives you ${(x+4)}^2=36$ and therefore $x=-4\pm \sqrt{36}$ so that $x=-10\vee x=2$.

$x=\frac{-8\pm \sqrt{144}}{2}$

$x^2+8x-20=(x+10)(x-2)$

$2x^2-x+1=10-3x$ gives you $2x^2+2x-9=0$ and $x=\frac{-2\pm \sqrt{76})}{2}$.

$x<\mathrm{-2,679}\vee x>\mathrm{1,679}$

$x^2-3x-13=0$ gives you $x=\frac{3\pm \sqrt{61})}{2}$.

$x^2+30x+3=0$ gives you $x=\frac{-30\pm \sqrt{888}}{2}$.

$2x^2-6x=0$ gives you $2x(x-3)=0$ and $x=0\vee x=3$.

$2x^2-12x+18=0$ gives you $x^2-6x+9=0$ or ${(x-3)}^2=0$ so that $x=3$.

$x^2-5x+10=0$ with $D=-15$, there are no solutions.

$x^2-x-12=0$ gives you $(x-4)(x+3)=0$ and therefore $x=4\vee x=-3$.

$x^2=60$ gives you $x=\pm \sqrt{60}$.

$\frac{1}{3}{x}^2=4$ gives you $x^2=12$ and therefore $x=\pm \sqrt{12}$.

$5x^2-x+3=0$ and $D=-59$, no solutions.

$f\left(x\right)=2{(x+1\frac{1}{2})}^2-2\frac{1}{2}$ gives you vertex $(-1\frac{1}{2},-2\frac{1}{2})$.

$D=36-8p^2=0$ gives you $p=\pm \sqrt{\mathrm{4,5}}$.
Also, the graph of $f$ is a straight line if $p=0$. Then it also has precisely one point in common with the $x$ -axis.

If $D>0$ and therefore for $p<\mathrm{-}\sqrt{\mathrm{4,5}}\vee p>\sqrt{(4,5)}$

$p{x}^2+6x+2p=6-x$ gives you $p{x}^2+7x+2p-6=0$.
$D=0$ gives you $49-4p(2p-6)=0$ and this means $p=\frac{24\pm \sqrt{2144}}{16}$.
There is also one intersection point if $p=0$.