find the solution set in a+bi form: 4x^2+7x+6=0?

4x² + 7x + 6 = 0
x² + 7/4x = - 3/2
x² + 7/8x = - 3/2 + (7/8)²
x² + 7/8x = - 96/64 + 56/64
(x + 7/8)² = - 40/64
x + 7/8 = 40i/8

x = 40i/8 - 7/8, x = (40i - 7)/8
x = - 40i/8 - 7/8, x = - (40i + 7)/8

Answer: x = (40i - 7)/8, - (40i + 7)/8

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Use the well known quadratic solution formula to fins

4x² + 7x + 6 = 0
x = {-7 ± √[7² - 4(4)(6)]}/2(4)
x = {-7 ± √-47}/8
x = (- 7 ± i√47)/8
x = - 7/8 ± i(√47)/8
x = - 7/8 ± (⅛√47)i

(-7+i sqrt47)/8 , (-7 - i sqrt47)/8

x = [ - 7 ± √ (49 - 96 ) ] / 8
x = [ - 7 ± √ (- 47 ) ] / 8
x = [ - 7 ± i √47 ] / 8

4x^2 + 7x + 6 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 4
b = 7
c = 6

x = [-7 ±√(49 - 144)]/8
x = [-7 ±√(-95)]/8

∴ x = [-7 ±√(-95)]/8

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isnt it holiday?

using the discriminant:

D= b^2 - 4ac
=49 - 96
= - 47 >>> it has no real roots!