1) 2m^2+10m=-12
2)(a-2)(a-3)=6
3) 3/x+5/3x=1
4)3/t-3=5/t+4
5) m=r/1+rt for r
6) 3(square root) x+5=12
7)(square root) x^2+3x+6=4
solve by factoring
8) x^2-2x-24=0
9)2x^2+7x=4
solve by the quadratic formula
10) x^2-5x-6=0

Hi cleanncrazi,

Here are the answers to your questions:

1. 2m^2+10m=-12
   2m^2+10m+12=0
   m^2+5m+6=0
   (m+3)(m+2)=0
   m=-3 and m=-2

2. (a-2)(a-3)=6
   a^2-3a-2a+6=6
   a^2-5a=0
   a(a-5)=0
   a-5=0
   a=5

3. 3/x+5/3x=1
   3/x+5/3x-1=0
   3(3x)+5(x)-1(x)(3x)=0
   9x+5x-3x^2=0
   -3x^2+14x=0
   x(-3x+14)=0
   -3x=-14
   x=14/3

4. 3/t-3=5/t+4
   (3-3t)/t=(5+4t)/t
   0= 5+4t-3+3t
   0=2+7t
   -2=7t
   -2/7=t

5. m=r/1+rt for r (assuming this is meant to be m=r/(1+rt)
   m+rtm=r
   m=r-rtm
   m=r(1-tm)
   m/(1-tm)=r
      
6. 3sqrtx+5=12
   9(x+5)=144
   x+5=16
   x=11

7. (square root) x^2+3x+6=4
   x^2+3x+6=4^2
   x^2+3x+6=16
   x^2+3x-10=0
   (x+5)(x-2)=0
   x=-5, x=2

8. x^2-2x-24=0
   (x+6)(x+4)=0
   x=6, x=4

9. 2x^2+7x=4
   2x^2+7x-4=0
   (x+4)(2x-1)=0
   x=-4, x=1/2

10. x^2-5x-6=0
    x=-[(-5)+-sqrt[(-5)^2-4(1)(6)]]/2
    x=[5+-sqrt(25+24)]/2
    x=[5+-sqrt(49)]/2
    
    x1=[5+sqrt(49)]/2
    x1=(5+7)/2
    x1=6

    x2=[5-sqrt(49)]/2
    x2=(5-7)/2
    x2=-1
	
Questions 1, 8, and 9 were done using the simple factoring method as
demonstrated at the following link:

Purplemath - Factoring Quadratics Lesson - 1
http://www.purplemath.com/modules/factquad.htm

I hope this helps! If you have any questions please feel free to ask
for clarification.

Regards,
Deadlychiapet

Very clear and concise work, thank you.