What are the solutions to the equation n^2-22n+102=0 ?

The solutions to the equation n^2-22n+102=0 are n=11+sqrt(19),n=11-sqrt(19)

Find the zeros of n^2-22n+102=0

The zeros of n^2-22n+102=0 are n=11+sqrt(19),n=11-sqrt(19)

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n^{2} - 22 n + 102 = 0

n = 11 + 19, n = 11 - 19

Decimal Notation

n = 15.35889 \dots, n = 6.64110 \dots

Solution steps

n^{2} - 22 n + 102 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 1 \cdot 102}{2 \cdot 1}

(- 22)^{2} - 4 \cdot 1 \cdot 102 = 219

n_{1, 2} = \frac{- ( - 22 ) \pm 2 19}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 22 ) + 2 19}{2 \cdot 1}, n_{2} = \frac{- ( - 22 ) - 2 19}{2 \cdot 1}

n = \frac{- ( - 22 ) + 2 19}{2 \cdot 1} : 11 + 19

n = \frac{- ( - 22 ) - 2 19}{2 \cdot 1} : 11 - 19

The solutions to the quadratic equation are:

n = 11 + 19, n = 11 - 19

12 x^{2} - 520 x + 4000 = 0

x = 5 x^{2}

(x + 2)^{2} = 64

a^{2} = 9

v^{2} - v = 72

What are the solutions to the equation n^2-22n+102=0 ?
The solutions to the equation n^2-22n+102=0 are n=11+sqrt(19),n=11-sqrt(19)
Find the zeros of n^2-22n+102=0
The zeros of n^2-22n+102=0 are n=11+sqrt(19),n=11-sqrt(19)