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

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

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

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

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n^{2} - 3 n = - 2

n = 2, n = 1

Solution steps

n^{2} - 3 n = - 2

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 2, n = 1

- 4 + 1

\frac{- 3}{2}

6 - x \geq 5 - 4 (x - 1)

49 r^{2} - 4 = 0

\frac{( x + 3 ) ^{2}}{4} = 9

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