What are the solutions to the equation (7x+4)^2=48 ?

The solutions to the equation (7x+4)^2=48 are x=(4sqrt(3)-4)/7 ,x=(-4sqrt(3)-4)/7

Find the zeros of (7x+4)^2=48

The zeros of (7x+4)^2=48 are x=(4sqrt(3)-4)/7 ,x=(-4sqrt(3)-4)/7

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(7 x + 4)^{2} = 48

x = \frac{4 3 - 4}{7}, x = \frac{- 4 3 - 4}{7}

Decimal Notation

x = 0.41831 \dots, x = - 1.56117 \dots

Solution steps

(7 x + 4)^{2} = 48

For

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

the solutions are

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

Solve

7 x + 4 = 48 : x = \frac{4 3 - 4}{7}

Solve

7 x + 4 = - 48 : x = \frac{- 4 3 - 4}{7}

The solutions to the quadratic equation are:

x = \frac{4 3 - 4}{7}, x = \frac{- 4 3 - 4}{7}

3 x^{2} + 5 x = 7 (x^{2} + x)

5 x^{2} + 40 x - 410 = 0

y, y^{2} - y - x^{2} = 0

2 (x - 1)^{2} - (x + 4) (x - 4) = 19 - x

2 x^{2} - 3 x - 10

What are the solutions to the equation (7x+4)^2=48 ?
The solutions to the equation (7x+4)^2=48 are x=(4sqrt(3)-4)/7 ,x=(-4sqrt(3)-4)/7
Find the zeros of (7x+4)^2=48
The zeros of (7x+4)^2=48 are x=(4sqrt(3)-4)/7 ,x=(-4sqrt(3)-4)/7