What are the solutions to the equation n^2-3n-28=0 ?

The solutions to the equation n^2-3n-28=0 are n=7,n=-4

Find the zeros of n^2-3n-28=0

The zeros of n^2-3n-28=0 are n=7,n=-4

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n^{2} - 3 n - 28 = 0

n = 7, n = - 4

Solution steps

n^{2} - 3 n - 28 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 7, n = - 4

16 x^{4} - 49 x^{2} y^{2} + 25 x y

- \frac{5}{6} - \frac{1}{6}

(- \frac{11}{6}) + m = - \frac{2}{9}

2 b^{3} - 6 b^{2}

x^{2} + (4)^{2} = (13)^{2}

What are the solutions to the equation n^2-3n-28=0 ?
The solutions to the equation n^2-3n-28=0 are n=7,n=-4
Find the zeros of n^2-3n-28=0
The zeros of n^2-3n-28=0 are n=7,n=-4