What are the solutions to the equation 3(4x+3)^2+1=16 ?

The solutions to the equation 3(4x+3)^2+1=16 are x=(sqrt(5)-3)/4 ,x=(-sqrt(5)-3)/4

Find the zeros of 3(4x+3)^2+1=16

The zeros of 3(4x+3)^2+1=16 are x=(sqrt(5)-3)/4 ,x=(-sqrt(5)-3)/4

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3 (4 x + 3)^{2} + 1 = 16

x = \frac{5 - 3}{4}, x = \frac{- 5 - 3}{4}

Decimal Notation

x = - 0.19098 \dots, x = - 1.30901 \dots

Solution steps

3 (4 x + 3)^{2} + 1 = 16

Move

1

to the right side

3 (4 x + 3)^{2} = 15

Divide both sides by

3

(4 x + 3)^{2} = 5

For

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

the solutions are

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

Solve

4 x + 3 = 5 : x = \frac{5 - 3}{4}

Solve

4 x + 3 = - 5 : x = \frac{- 5 - 3}{4}

The solutions to the quadratic equation are:

x = \frac{5 - 3}{4}, x = \frac{- 5 - 3}{4}

\frac{x}{3} x + 10 = 15

5 x^{2} - 8 x + 25 = 0

x^{2} - 8 x + 12 = 0

x^{2} + 1 = 0

(x - 4) (1 + x) = 0

What are the solutions to the equation 3(4x+3)^2+1=16 ?
The solutions to the equation 3(4x+3)^2+1=16 are x=(sqrt(5)-3)/4 ,x=(-sqrt(5)-3)/4
Find the zeros of 3(4x+3)^2+1=16
The zeros of 3(4x+3)^2+1=16 are x=(sqrt(5)-3)/4 ,x=(-sqrt(5)-3)/4