What are the solutions to the equation 4(6x-1)^2-5=223 ?

The solutions to the equation 4(6x-1)^2-5=223 are x=(sqrt(57)+1)/6 ,x=(-sqrt(57)+1)/6

Find the zeros of 4(6x-1)^2-5=223

The zeros of 4(6x-1)^2-5=223 are x=(sqrt(57)+1)/6 ,x=(-sqrt(57)+1)/6

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4 (6 x - 1)^{2} - 5 = 223

x = \frac{57 + 1}{6}, x = \frac{- 57 + 1}{6}

Decimal Notation

x = 1.42497 \dots, x = - 1.09163 \dots

Solution steps

4 (6 x - 1)^{2} - 5 = 223

Move

5

to the right side

4 (6 x - 1)^{2} = 228

Divide both sides by

4

(6 x - 1)^{2} = 57

For

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

the solutions are

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

Solve

6 x - 1 = 57 : x = \frac{57 + 1}{6}

Solve

6 x - 1 = - 57 : x = \frac{- 57 + 1}{6}

The solutions to the quadratic equation are:

x = \frac{57 + 1}{6}, x = \frac{- 57 + 1}{6}

\frac{1}{16} - 1

30 x^{3} y + 35 x^{2} y^{2}

x + 23 = 9 x - 17

x^{2} - 16 x = - 63

x^{2} = 5 x - 10

What are the solutions to the equation 4(6x-1)^2-5=223 ?
The solutions to the equation 4(6x-1)^2-5=223 are x=(sqrt(57)+1)/6 ,x=(-sqrt(57)+1)/6
Find the zeros of 4(6x-1)^2-5=223
The zeros of 4(6x-1)^2-5=223 are x=(sqrt(57)+1)/6 ,x=(-sqrt(57)+1)/6