Simple algebra problem?? help

How does

8^x = 4^(x^2-3)

so the answer is x = (3 +- sqrt(57)) / 4

how does this answer come out?? please help show steps please

2^(3x) = 2^[2(x^2 - 3)]
3x = 2x^2 - 6
2x^2 - 3x - 6 = 0

thus x = [3 +_ sqrt (9 + 48) ]/4
x = (3 +- sqrt(57)) / 4

8^x = (2^3)^x = 2^(3x)
4^(x² - 3) = {(2^2)}^(x²- 3) = 2^{2(x²-3)} = 2^(2x²- 6)
8^x = 4^(x² - 3)
=> 2^(3x) = 2^(2x²- 6)
=> 3x = (2x² - 6)
=> 2x² - 3x - 6 = 0
=>(4*2)*2x² - (4*2)3x - (4*2)*6 = 0
=> (4x)² - 2*(4x)*3 = 4*2*6
=> (4x)² - 2*(4x)*3 + 3² = 48 + 3²
=> (4x - 3)² = 57
=> 4x - 3 = ± √57
=> 4x = 3 ± √57
=> x = (3 ± √57) / 4
You say simple Algebra problem and you need help ? Wonderful !
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x log 8 = (x² - 3) log 4
3x = 2 (x² - 3)_________taking logs to base 2
2x² - 3x - 6 = 0
x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a
x = [ 3 ± √ ( 9 + 48 ) ] / 4
x = [ 3 ± √57 ] / 4

8^x = 4^(x^2-3)


2^(3x) =2^[2(x²-3)]

3x=2x²-6

2x²-3x-6=0

Δ= 57

-b(+/-)√Δ
----------------
....2.*a.


x'=3(+/-) √57
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........4......