4(1/2a^2+1^2)-5=31

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Solution for 4(1/2a^2+1^2)-5=31 equation: 


a in (-oo:+oo)

4*((1/2)*a^2+1^2)-5 = 31 // - 31

4*((1/2)*a^2+1^2)-31-5 = 0

4*((1/2)*a^2+1)-31-5 = 0

4*(1/2*a^2+1)-31-5 = 0

4*(1/2*a^2+1)-31-5 = 0

2*a^2-27-5 = 0

2*a^2-32 = 0

2*a^2-32 = 0

2*a^2 = 32 // : 2

a^2 = 16

a^2 = 16 // ^ 1/2

abs(a) = 4

a = 4 or a = -4

a in { 4, -4 }

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