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

The solutions to the equation -n^2=3-4n are n=1,n=3

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

The zeros of -n^2=3-4n are n=1,n=3

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- n^{2} = 3 - 4 n

n = 1, n = 3

Solution steps

- n^{2} = 3 - 4 n

Move

4 n

to the left side

- n^{2} + 4 n = 3

Move

3

to the left side

- n^{2} + 4 n - 3 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 1) (- 3) = 2

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 1, n = 3

v^{2} - 9. 8^{2} = 2 \cdot 9.8 \cdot 19.6

(4 x - 1)^{2} + (- 2)^{3} = 2020

(k + 2) \cdot x^{2} - 13 x + 4 k - 10 = 0

(7 - 2 f)^{2} = 9

x, f = x^{2} - 3 x - 2

What are the solutions to the equation -n^2=3-4n ?
The solutions to the equation -n^2=3-4n are n=1,n=3
Find the zeros of -n^2=3-4n
The zeros of -n^2=3-4n are n=1,n=3