What are the solutions to the equation 3(x+1)^2-6=141 ?

The solutions to the equation 3(x+1)^2-6=141 are x=6,x=-8

Find the zeros of 3(x+1)^2-6=141

The zeros of 3(x+1)^2-6=141 are x=6,x=-8

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3 (x + 1)^{2} - 6 = 141

x = 6, x = - 8

Solution steps

3 (x + 1)^{2} - 6 = 141

Move

6

to the right side

3 (x + 1)^{2} = 147

Divide both sides by

3

(x + 1)^{2} = 49

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

x + 1 = 49 : x = 6

Solve

x + 1 = - 49 : x = - 8

The solutions to the quadratic equation are:

x = 6, x = - 8

3 m^{2} - 4 m - 4 = m

4 x^{2} + 100 = 2 x^{2} + 492

y, 36 x^{2} + 36 y^{2} + 48 x - 108 y + 97 = 0

y, x^{2} - x y - y^{2} = 1

x^{2} + 7 x + 3 = 3

What are the solutions to the equation 3(x+1)^2-6=141 ?
The solutions to the equation 3(x+1)^2-6=141 are x=6,x=-8
Find the zeros of 3(x+1)^2-6=141
The zeros of 3(x+1)^2-6=141 are x=6,x=-8