Can anyone solve this MATH PROBLEM?

Question

problem is

3(y-7) to power of 2 = -12

3(y-7)^2 = -12

Answer 1

3(y-7)² = -12
(y-7)² = -4
y - 7 = ±2i
y = 7 ± 2i

Answer 2

21 6 ^ -12 +2= 12
+
$12-

Answer 3

Last I checked there is no answer. If y=5 then -2 is squared but when -2 x -2 = +4 not -4. Any even number of + or - numbers always yield a + number. Since 3 is +3 and any thing you square is + then there is no answer

my bad I was thinking irrationally

Answer 4

ATTENTION CHRIS!!!

you are right there is no square root for any real negative numbers so to get the square root of a negative number you work with a greater set of numbers called the COMPLEX NUMBERS where

i^2 = -1 so i = sqrt(-1), where now ' i ' is an IMAGINARY UNIT


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3(y-7) ^2 = -12

take square roots on both sides

sq(3) *( y -7) = (+,- )sq(-12) = (+,- ) sq(-1) *sq(12) = (+,- ) i *sq(12)

y -7 = i * sq(12)/sq(3) = (+, - ) 2* i

y = 7 + 2i or y = 7 - 2i

Answer 5

There's something wrong with your question because:

3(y-7)^2=-12 so
3y^2+147=-12 so
3y^2=-161 so
y^2=-53 2/3 so
y= radical -53 2/3 but you can't have a negative radical (square root)

sorry

a lot of these people are wrong because you CAN'T EVER EVER EVER have a radical (square root) of a negative (-) number EVER!

Answer 6

3(y-7)^2 = -12

divide by 3

(y - 7)^2 = -4

y- 7 = +/-sqrt(-4)

y - 7 = +/- 2i

y = 7 + 2i or 7 - 2i

Answer 7

3(y-7)^2 = -12

(y-7)^2 = -12/3

(y-7)^2 = - 4 -------> -1(2)(2) square root = 2

(y-7) = - 2

y = -2 + 7

y = 5

Answer 8

(y-7)^2=-4

Square Root; it becomes imaginary...

y-7=2i
y=2i+7