What are the solutions to the equation -2n^2+2n+53=-7 ?

The solutions to the equation -2n^2+2n+53=-7 are n=-5,n=6

Find the zeros of -2n^2+2n+53=-7

The zeros of -2n^2+2n+53=-7 are n=-5,n=6

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- 2 n^{2} + 2 n + 53 = - 7

n = - 5, n = 6

Solution steps

- 2 n^{2} + 2 n + 53 = - 7

Move

7

to the left side

- 2 n^{2} + 2 n + 60 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 60}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 60 = 22

n_{1, 2} = \frac{- 2 \pm 22}{2 ( - 2 )}

Separate the solutions

n_{1} = \frac{- 2 + 22}{2 ( - 2 )}, n_{2} = \frac{- 2 - 22}{2 ( - 2 )}

n = \frac{- 2 + 22}{2 ( - 2 )} : - 5

n = \frac{- 2 - 22}{2 ( - 2 )} : 6

The solutions to the quadratic equation are:

n = - 5, n = 6

2 x^{2} + 4 x + 6

3 x^{2} = - 8 x - 4

x^{2} + 7 = 18

x, f = (x - k)^{2} + 5

x^{2} - 76.7 x + 259 = 0

What are the solutions to the equation -2n^2+2n+53=-7 ?
The solutions to the equation -2n^2+2n+53=-7 are n=-5,n=6
Find the zeros of -2n^2+2n+53=-7
The zeros of -2n^2+2n+53=-7 are n=-5,n=6