What are the solutions to the equation n^2+32=12n ?

The solutions to the equation n^2+32=12n are n=8,n=4

Find the zeros of n^2+32=12n

The zeros of n^2+32=12n are n=8,n=4

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n^{2} + 32 = 12 n

n = 8, n = 4

Solution steps

n^{2} + 32 = 12 n

Move

12 n

to the left side

n^{2} + 32 - 12 n = 0

Write in the standard form

a x^{2} + b x + c = 0

n^{2} - 12 n + 32 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

n_{1, 2} = \frac{- ( - 12 ) \pm 4}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, n_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

n = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

n = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

n = 8, n = 4

3 x^{2} - 5 x - 50 = 0

3 m^{2} - 19 m - 14 = 0

x^{2} + 12 x - 189 = 0

3 x^{2} + 18 x - 48 = 0

x, (x - 14)^{2} \cdot 1 0^{- 15} = 3

What are the solutions to the equation n^2+32=12n ?
The solutions to the equation n^2+32=12n are n=8,n=4
Find the zeros of n^2+32=12n
The zeros of n^2+32=12n are n=8,n=4