What are the solutions to the equation 2n-4=n^2-3n-14 ?

The solutions to the equation 2n-4=n^2-3n-14 are n=(5+sqrt(65))/2 ,n=(5-sqrt(65))/2

Find the zeros of 2n-4=n^2-3n-14

The zeros of 2n-4=n^2-3n-14 are n=(5+sqrt(65))/2 ,n=(5-sqrt(65))/2

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2 n - 4 = n^{2} - 3 n - 14

n = \frac{5 + 65}{2}, n = \frac{5 - 65}{2}

Decimal Notation

n = 6.53112 \dots, n = - 1.53112 \dots

Solution steps

2 n - 4 = n^{2} - 3 n - 14

Switch sides

n^{2} - 3 n - 14 = 2 n - 4

Move

4

to the left side

n^{2} - 3 n - 10 = 2 n

Move

2 n

to the left side

n^{2} - 5 n - 10 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 10) = 65

n_{1, 2} = \frac{- ( - 5 ) \pm 65}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 5 ) + 65}{2 \cdot 1}, n_{2} = \frac{- ( - 5 ) - 65}{2 \cdot 1}

n = \frac{- ( - 5 ) + 65}{2 \cdot 1} : \frac{5 + 65}{2}

n = \frac{- ( - 5 ) - 65}{2 \cdot 1} : \frac{5 - 65}{2}

The solutions to the quadratic equation are:

n = \frac{5 + 65}{2}, n = \frac{5 - 65}{2}

n^{2} - 7 n + 10 = 0

- x^{2} - 6 x + 2

- x^{2} - 6 x + 1

(7 + x) (9 + x) = 92

\frac{1}{2} x^{2} + 2 x + 2 = 0

What are the solutions to the equation 2n-4=n^2-3n-14 ?
The solutions to the equation 2n-4=n^2-3n-14 are n=(5+sqrt(65))/2 ,n=(5-sqrt(65))/2
Find the zeros of 2n-4=n^2-3n-14
The zeros of 2n-4=n^2-3n-14 are n=(5+sqrt(65))/2 ,n=(5-sqrt(65))/2