Question

What values of b satisfy 3(2b + 3)2 = 36?

b = StartFraction negative 3 + 2 StartRoot 3 EndRoot Over 2 EndFraction and StartFraction negative 3 minus 2 StartRoot 3 EndRoot Over 2 EndFraction
b = StartFraction negative 3 + 2 StartRoot 3 EndRoot Over 2 EndFraction and StartFraction negative 3 minus 2 StartRoot 3 EndRoot Over 2 EndFraction
b = StartFraction 3 Over 2 EndFraction and StartFraction negative 9 Over 2 EndFraction
b = Start Fraction 9 Over 2 EndFraction and

Answer 1

`b = (-3\pm2√(3))/(2)`

Explanation:

`3(2b+3)^(2) = 36 \\\\(2b+3)^(2) = 12 \\\\2b+3 = \pm√(12) = \pm2√(3) \\\\b = (-3\pm2√(3))/(2)`