Solve each one:

-x^2+4x-13 = 0

x^2 - 4x + 13 = 0      Use quadratic formula  

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)       a = 1  b  = -4   c = 13

Sustituting yields       4 +- sqrt( 16- 4*1*13)   /  2     =    2 +-sqrt (-36)/2    =  2 +-  3i

Think you can try the others?

b )  x^2 + 41 = -8x

x^2 + 8x  =  -41      complete the square on x

x^2 + 8x + 16  =   -41 + 16

(x + 4)^2 =   - 25      take both roots

x + 4 =  ±√-25

x + 4 = ± 5i      subtract 4 from both sides

x =   - 4 ± 5i

(c)

-9x^2 = 30x + 25   rearrange as

9x^2 + 30x + 25 = 0        we can factor this    !!!

(3x + 5) ( 3x + 5) = 0

Setting either factor to 0 and solving for x produces     x =  - 5/3

Lastly ...

5 - 7x^2 = 3x + 4  rearrange as

7x^2  + 3x + 4 - 5 = 0  simplify

7x^2 + 3x - 1 = 0

Quad formula

 x =  (  -3 ±√  [ 3^2 - 4 (7) (-1) ]  )  / [ 2 * 7 ] 

x = ( - 3 ±√ 37 ) /  14