need help!1 how do i solve these? thank you!11?

Question

need help!1 how do i solve these? thank you!11

Find/Solve for X:

1. x²+4x=7
2. -12x²+25x +12=0

Convert to a + bi form:
1. (square root of -12) - (square root of -48)

Edit: Can someone work out problem two plz! the 3rd response i believe under the square root it is 1201 not 49??

Answer 1

1.
x^2 + 4x - 7 = 0
x^2 + 4x + 4 - 11 = 0
(x + 2)^2 - 11 = 0
(x + 2)^2 = 11
x + 2 = square root of 11
x = square root of 11 - 2

Answer 2

1. x² + 4x -7 = 0

quadratic formula:
-b ± √(b² -4ac)
/2a

-4 ± √(4² -4*1*-7)
/2*1

-4 ± √(16 + 28)
/2

-4 ± √44
/2

-4 ± √4√11
/2

x = -2 ± 2√11

2. -12x² + 25x + 12 = 0

-b ± √(b² -4ac)
/2a

-25 ± √(25² -4*-12*12)
/2*-12

-25 ± √(625 -576)
/-24

-25 ± √49
/-24

-25 ± 7
/-24

-32/-24 or -12/-24

x = 1/2 or 4/3

1. √-12 - √-48
no real numbers,
a negative can't be under a square root

Answer 3

x^2+4x-7=0 so x= ((-4+-sqrt(44))/2 =-2+-sqrt(11)
Apply the same formula to 2)
sqrt(-12 ) =i*2*sqrt(3) and sqrt(-48) = i*4*sqrt(3)
so the difference is -2i*sqrt(3) =b*i and a=0