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1. What is the largest value of k such that the equation 6x-x^2=k has at least one real solution?

Rearrange as    x^2 - 6x + k = 0

If this has at least one real solution, the discriminant must be ≥ 0

So

(-6)^2 -   4(1)(k) ≥ 0

36 - 4k ≥ 0

36 ≥ 4k 

9 ≥ k

So .....the max value  of k that produces at least one real solution is when k = 9

2. 

move all the numbers on one side so that the right side is left with 1

x^2+6x-15=0

use the quadratic formula to find smallest x can be

you should get something like

x = -3-2sqrt(6)

2.Find the smallest real value of x such that x^2+6x+9=24.

Factor the left side

(x + 3)^2  = 24               take the negative root  [ since we want the smallest value ]

x + 3 = - √24       subtract 3 from both sides

x =  -√24 - 3  =    -2√6 - 3      and this is the smallest real value that makes the equation true

4.Find all real solutions to x^2+4=100x^2+20x+1.

Rearrange as

99x^2 + 20x - 3   =  0

Using the quad formula.....we have

( -20 ±√ [ 20^2 - 4 * 99 * -3 ]  ) / (2 * 99)

( -20 ±√ 1588) / 198

(-20 ± 2√397 )/198 =

(-10 ±√397 ) / 99   .....and these are the two solutions

3.

Similarly to question 2, you should simplify all the like terms first

x^2+12x+61=0

in ax^2+bx=c you get a=1 b=12 c=61

use the quadratic formula to find the roots of the equation

solve this and you should get the 2 roots

x= -6+5i , x= -6-5i