What are the solutions to the equation n-5n+12=3n^2 ?

The solutions to the equation n-5n+12=3n^2 are n=(2(sqrt(10)-1))/3 ,n=-(2(1+sqrt(10)))/3

Find the zeros of n-5n+12=3n^2

The zeros of n-5n+12=3n^2 are n=(2(sqrt(10)-1))/3 ,n=-(2(1+sqrt(10)))/3

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n - 5 n + 12 = 3 n^{2}

n = \frac{2 ( 10 - 1 )}{3}, n = - \frac{2 ( 1 + 10 )}{3}

Solution steps

n - 5 n + 12 = 3 n^{2}

Refine

- 4 n + 12 = 3 n^{2}

Switch sides

3 n^{2} = - 4 n + 12

Move

12

to the left side

3 n^{2} - 12 = - 4 n

Move

4 n

to the left side

3 n^{2} - 12 + 4 n = 0

Write in the standard form

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

3 n^{2} + 4 n - 12 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

4^{2} - 4 \cdot 3 (- 12) = 410

n_{1, 2} = \frac{- 4 \pm 4 10}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 4 + 4 10}{2 \cdot 3}, n_{2} = \frac{- 4 - 4 10}{2 \cdot 3}

n = \frac{- 4 + 4 10}{2 \cdot 3} : \frac{2 ( 10 - 1 )}{3}

n = \frac{- 4 - 4 10}{2 \cdot 3} : - \frac{2 ( 1 + 10 )}{3}

The solutions to the quadratic equation are:

n = \frac{2 ( 10 - 1 )}{3}, n = - \frac{2 ( 1 + 10 )}{3}

x^{2} + 8 x - 9 = 0

- 6 x^{2} + 7 x = 7 x - 4

- z^{2} - 7 z - 4 = - 2 z^{2}

x, f (x - 2) = - 3 x^{2} - 1

x, y (x^{2} - 1) = 2

What are the solutions to the equation n-5n+12=3n^2 ?
The solutions to the equation n-5n+12=3n^2 are n=(2(sqrt(10)-1))/3 ,n=-(2(1+sqrt(10)))/3
Find the zeros of n-5n+12=3n^2
The zeros of n-5n+12=3n^2 are n=(2(sqrt(10)-1))/3 ,n=-(2(1+sqrt(10)))/3