ssepeda98 ssepeda98

11-01-2016
Mathematics

contestada

6(y+4)^2-5=49
a). y = -2 or y = -6
b). y = -1 or y = -7
c). y = 0 or y = -4
d). NO REAL SOLUTIONS

Respuesta :

Аноним Аноним

11-01-2016

☾Let's solve this problem Step-By-Step!☽
☾Question you asked:
6(y+4)^2−5=49

☾Simplify both sides of the equation:
6y^2+48y+91=49

☾Subtract 49 from both sides:
6y^2+48y+91−49=49−49
6y^2+48y+42=0

☾Factor left side of equation:
6(y+1)(y+7)=0

☾Set factors equal to 0:
y+1=0 or y+7=0

☾Your answer is:
y=−1 or y=−7

☾I hope this helps!☽

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