1/4(2b-12)+b2=1/2b+22

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Solution for 1/4(2b-12)+b2=1/2b+22 equation:


b in (-oo:+oo)

(1/4)*(2*b-12)+b^2 = (1/2)*b+22 // - (1/2)*b+22

(1/4)*(2*b-12)+b^2-((1/2)*b)-22 = 0

(1/4)*(2*b-12)+b^2+(-1/2)*b-22 = 0

1/4*(2*b-12)+b^2-b/2-22 = 0

(1/4*2*(2*b-12))/2+(2*b^2)/2-b/2+(-22*2)/2 = 0

1/4*2*(2*b-12)+2*b^2-b-22*2 = 0

2*b^2+b-b-44-6 = 0

2*b^2-44-6 = 0

2*b^2-50 = 0

(2*b^2-50)/2 = 0

(2*b^2-50)/2 = 0 // * 2

2*b^2-50 = 0

2*b^2 = 50 // : 2

b^2 = 25

b^2 = 25 // ^ 1/2

abs(b) = 5

b = 5 or b = -5

b in { 5, -5 }

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