What are the solutions to the equation 4n^2-22=27 ?

The solutions to the equation 4n^2-22=27 are n= 7/2 ,n=-7/2

Find the zeros of 4n^2-22=27

The zeros of 4n^2-22=27 are n= 7/2 ,n=-7/2

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4 n^{2} - 22 = 27

n = \frac{7}{2}, n = - \frac{7}{2}

Decimal Notation

n = 3.5, n = - 3.5

Solution steps

4 n^{2} - 22 = 27

Move

22

to the right side

4 n^{2} = 49

Divide both sides by

4

n^{2} = \frac{49}{4}

For

x^{2} = f (a)

the solutions are

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

n = \frac{49}{4}, n = - \frac{49}{4}

\frac{49}{4} = \frac{7}{2}

- \frac{49}{4} = - \frac{7}{2}

n = \frac{7}{2}, n = - \frac{7}{2}

20 r^{2} + 37 r + 15

16 y^{2} + 8 y - 15

(m - 3) (m - 4) = 20

3 (2 + r) > 15 - 2 r

5!

What are the solutions to the equation 4n^2-22=27 ?
The solutions to the equation 4n^2-22=27 are n= 7/2 ,n=-7/2
Find the zeros of 4n^2-22=27
The zeros of 4n^2-22=27 are n= 7/2 ,n=-7/2