What are the solutions to the equation n^2-4n=11 ?

The solutions to the equation n^2-4n=11 are n=2+sqrt(15),n=2-sqrt(15)

Find the zeros of n^2-4n=11

The zeros of n^2-4n=11 are n=2+sqrt(15),n=2-sqrt(15)

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n^{2} - 4 n = 11

n = 2 + 15, n = 2 - 15

Solution steps

n^{2} - 4 n = 11

Move

11

to the left side

n^{2} - 4 n - 11 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 11) = 215

n_{1, 2} = \frac{- ( - 4 ) \pm 2 15}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 4 ) + 2 15}{2 \cdot 1}, n_{2} = \frac{- ( - 4 ) - 2 15}{2 \cdot 1}

n = \frac{- ( - 4 ) + 2 15}{2 \cdot 1} : 2 + 15

n = \frac{- ( - 4 ) - 2 15}{2 \cdot 1} : 2 - 15

The solutions to the quadratic equation are:

n = 2 + 15, n = 2 - 15

x^{2} - 7 x - 34 = 0

x^{2} + 289 = 0

9 x^{2} + 7 x - 56 = 0

5 n^{2} + 19 n + 12 = 0

19 x^{2} = 0

What are the solutions to the equation n^2-4n=11 ?
The solutions to the equation n^2-4n=11 are n=2+sqrt(15),n=2-sqrt(15)
Find the zeros of n^2-4n=11
The zeros of n^2-4n=11 are n=2+sqrt(15),n=2-sqrt(15)