What are the solutions to the equation 98z^2+39=41 ?

The solutions to the equation 98z^2+39=41 are z= 1/7 ,z=-1/7

Find the zeros of 98z^2+39=41

The zeros of 98z^2+39=41 are z= 1/7 ,z=-1/7

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98 z^{2} + 39 = 41

z = \frac{1}{7}, z = - \frac{1}{7}

Decimal Notation

z = 0.14285 \dots, z = - 0.14285 \dots

Solution steps

98 z^{2} + 39 = 41

Move

39

to the right side

98 z^{2} = 2

Divide both sides by

98

z^{2} = \frac{1}{49}

For

x^{2} = f (a)

the solutions are

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

z = \frac{1}{49}, z = - \frac{1}{49}

\frac{1}{49} = \frac{1}{7}

- \frac{1}{49} = - \frac{1}{7}

z = \frac{1}{7}, z = - \frac{1}{7}

4 a^{3} - 36 a

∣ 2 x ∣ = ∣ x + 3 ∣

\frac{2 x ^{3} y ^{7}}{4 x ^{3} y ^{3}}

\frac{3}{1000} (x - 100)^{2} + 10 = 30

1 0^{2} + x^{2} = 1 2^{2}

What are the solutions to the equation 98z^2+39=41 ?
The solutions to the equation 98z^2+39=41 are z= 1/7 ,z=-1/7
Find the zeros of 98z^2+39=41
The zeros of 98z^2+39=41 are z= 1/7 ,z=-1/7