What are the solutions to the equation 6(5x+x)^2-42=0 ?

The solutions to the equation 6(5x+x)^2-42=0 are x=(sqrt(7))/6 ,x=-(sqrt(7))/6

Find the zeros of 6(5x+x)^2-42=0

The zeros of 6(5x+x)^2-42=0 are x=(sqrt(7))/6 ,x=-(sqrt(7))/6

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6 (5 x + x)^{2} - 42 = 0

x = \frac{7}{6}, x = - \frac{7}{6}

Decimal Notation

x = 0.44095 \dots, x = - 0.44095 \dots

Solution steps

6 (5 x + x)^{2} - 42 = 0

Move

42

to the right side

6 (5 x + x)^{2} = 42

Divide both sides by

6

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

For

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

the solutions are

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

Solve

5 x + x = 7 : x = \frac{7}{6}

Solve

5 x + x = - 7 : x = - \frac{7}{6}

The solutions to the quadratic equation are:

x = \frac{7}{6}, x = - \frac{7}{6}

7 x = 3 x^{2}

a^{2} = 225

2 x^{2} + 6 x - 8

x, 10 x^{2} + 10 y^{2} = 0

3 x^{2} = 5 - 14 x

What are the solutions to the equation 6(5x+x)^2-42=0 ?
The solutions to the equation 6(5x+x)^2-42=0 are x=(sqrt(7))/6 ,x=-(sqrt(7))/6
Find the zeros of 6(5x+x)^2-42=0
The zeros of 6(5x+x)^2-42=0 are x=(sqrt(7))/6 ,x=-(sqrt(7))/6