Question 362021
these are mostly quadratic equations(1,2,5,6,7,8,9,10) which are solved the same way you just need to do the calculations i will do one equation for u below:
#1
{{{r^3+r^2-6r}}}
{{{r(r^2+r-6)}}}
{{{r(r^2+r-6)}}} +3 and -2 are two numbers that multiply to get -6 and add to get +1, if you cant do it this way use the quadratic formula to solve.
{{{r((r-2)(r+3))}}}
{{{r-2=0}}}={{{2}}}
{{{r+3=0}}}={{{-3}}}
#3
{{{81m^4-16}}}
{{{(9m^2-4)(9m^2-4)}}}
{{{((3m-2)(3m+2))((3m-2)(3m+2))}}}
{{{3m=2}}}={{{2/3}}}
{{{3m=-2}}}={{{-2/3}}}
#4 Im not sure what the question is asking for you to do, cause u cant find what either equals with 2 variables makes near unlimited possible solutions
#10
{{{8a^2+23a-3}}}
*[invoke quadratic "a", 8, 23, -3]