What are the solutions to the equation 1.5=(4/81)(x-3)^2+1 ?

The solutions to the equation 1.5=(4/81)(x-3)^2+1 are x=6.18198…,x=-0.18198…

Find the zeros of 1.5=(4/81)(x-3)^2+1

The zeros of 1.5=(4/81)(x-3)^2+1 are x=6.18198…,x=-0.18198…

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1.5 = (\frac{4}{81}) (x - 3)^{2} + 1

x = 6.18198 \dots, x = - 0.18198 \dots

Solution steps

1.5 = (\frac{4}{81}) (x - 3)^{2} + 1

Switch sides

(\frac{4}{81}) (x - 3)^{2} + 1 = 1.5

Move

1

to the right side

\frac{4}{81} (x - 3)^{2} = 0.5

Multiply both sides by

81

4 (x - 3)^{2} = 40.5

Divide both sides by

4

(x - 3)^{2} = 10.125

For

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

the solutions are

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

Solve

x - 3 = 10.125 : x = 6.18198 \dots

Solve

x - 3 = - 10.125 : x = - 0.18198 \dots

The solutions to the quadratic equation are:

x = 6.18198 \dots, x = - 0.18198 \dots

3 x^{2} + 2 x - 1 = 2

1 5^{2} = x^{2} + 7. 5^{2}

p, 10 - p^{2} = 3 p

X^{2} - 2 X = - X + 6

- 5 x^{2} - 40 x - 86

What are the solutions to the equation 1.5=(4/81)(x-3)^2+1 ?
The solutions to the equation 1.5=(4/81)(x-3)^2+1 are x=6.18198…,x=-0.18198…
Find the zeros of 1.5=(4/81)(x-3)^2+1
The zeros of 1.5=(4/81)(x-3)^2+1 are x=6.18198…,x=-0.18198…