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

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

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

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

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

x = 3, x = - 3

Decimal Notation

x = 1.73205 \dots, x = - 1.73205 \dots

Solution steps

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

Move

5

to the right side

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

Divide both sides by

4

x^{2} - 4 = - 1

Move

4

to the right side

x^{2} = 3

For

x^{2} = f (a)

the solutions are

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

x = 3, x = - 3

\frac{10}{6}

6 x^{2} - 10 x = 0

(\frac{4 x ^{- 3} y ^{4}}{8 x ^{2} y ^{- 2}})^{- 2}

- 2 x^{2} + 98 = 0

5 ∣ x + 9 ∣ + 2 < 17

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