What are the solutions to the equation n^2-9n+10=0 ?

The solutions to the equation n^2-9n+10=0 are n=(9+sqrt(41))/2 ,n=(9-sqrt(41))/2

Find the zeros of n^2-9n+10=0

The zeros of n^2-9n+10=0 are n=(9+sqrt(41))/2 ,n=(9-sqrt(41))/2

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n^{2} - 9 n + 10 = 0

n = \frac{9 + 41}{2}, n = \frac{9 - 41}{2}

Decimal Notation

n = 7.70156 \dots, n = 1.29843 \dots

Solution steps

n^{2} - 9 n + 10 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 10 = 41

n_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 1}, n_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 1}

n = \frac{- ( - 9 ) + 41}{2 \cdot 1} : \frac{9 + 41}{2}

n = \frac{- ( - 9 ) - 41}{2 \cdot 1} : \frac{9 - 41}{2}

The solutions to the quadratic equation are:

n = \frac{9 + 41}{2}, n = \frac{9 - 41}{2}

A = 2 π r^{2} + 2 π r h, r

x, f = x^{2} + m x + m

x (2 x - 5) = - 3

(x + \frac{1}{2}) (x - 2) = 0

4 x^{2} + 6 = 25 x

What are the solutions to the equation n^2-9n+10=0 ?
The solutions to the equation n^2-9n+10=0 are n=(9+sqrt(41))/2 ,n=(9-sqrt(41))/2
Find the zeros of n^2-9n+10=0
The zeros of n^2-9n+10=0 are n=(9+sqrt(41))/2 ,n=(9-sqrt(41))/2