2x+(4x^2+7x+3)^1/2

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Solution for 2x+(4x^2+7x+3)^1/2 equation: 


x in (-oo:+oo)

((4*x^2+7*x+3)^1)/2+2*x = 0

(4*x^2+7*x+3)/2+2*x = 0

(4*x^2+7*x+3)/2+(2*2*x)/2 = 0

4*x^2+7*x+2*2*x+3 = 0

4*x^2+11*x+3 = 0

4*x^2+11*x+3 = 0

4*x^2+11*x+3 = 0

DELTA = 11^2-(3*4*4)

DELTA = 73

DELTA > 0

x = (73^(1/2)-11)/(2*4) or x = (-73^(1/2)-11)/(2*4)

x = (73^(1/2)-11)/8 or x = (-(73^(1/2)+11))/8

(x+(73^(1/2)+11)/8)*(x-((73^(1/2)-11)/8)) = 0

((x+(73^(1/2)+11)/8)*(x-((73^(1/2)-11)/8)))/2 = 0

((x+(73^(1/2)+11)/8)*(x-((73^(1/2)-11)/8)))/2 = 0 // * 2

(x+(73^(1/2)+11)/8)*(x-((73^(1/2)-11)/8)) = 0

( x+(73^(1/2)+11)/8 )

x+(73^(1/2)+11)/8 = 0 // - (73^(1/2)+11)/8

x = -((73^(1/2)+11)/8)

( x-((73^(1/2)-11)/8) )

x-((73^(1/2)-11)/8) = 0 // + (73^(1/2)-11)/8

x = (73^(1/2)-11)/8

x in { -((73^(1/2)+11)/8), (73^(1/2)-11)/8 }

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