What are the solutions to the equation n^2-11n+28=0 ?

The solutions to the equation n^2-11n+28=0 are n=7,n=4

Find the zeros of n^2-11n+28=0

The zeros of n^2-11n+28=0 are n=7,n=4

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n^{2} - 11 n + 28 = 0

n = 7, n = 4

Solution steps

n^{2} - 11 n + 28 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 11)^{2} - 4 \cdot 1 \cdot 28 = 3

n_{1, 2} = \frac{- ( - 11 ) \pm 3}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 11 ) + 3}{2 \cdot 1}, n_{2} = \frac{- ( - 11 ) - 3}{2 \cdot 1}

n = \frac{- ( - 11 ) + 3}{2 \cdot 1} : 7

n = \frac{- ( - 11 ) - 3}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

n = 7, n = 4

x^{2} + 13 x - 140 = 0

2 k (k - 4) = 1

x, z^{2} = x^{2} + y^{2}

x^{2} - 6 x - 20 = 20

4 x^{2} - 3 x - 12 = 3 x^{2} - 6 x - 14

What are the solutions to the equation n^2-11n+28=0 ?
The solutions to the equation n^2-11n+28=0 are n=7,n=4
Find the zeros of n^2-11n+28=0
The zeros of n^2-11n+28=0 are n=7,n=4