What are the solutions to the equation 6=n^2+3n-12 ?

The solutions to the equation 6=n^2+3n-12 are n=3,n=-6

Find the zeros of 6=n^2+3n-12

The zeros of 6=n^2+3n-12 are n=3,n=-6

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

n = 3, n = - 6

Solution steps

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

Switch sides

n^{2} + 3 n - 12 = 6

Move

6

to the left side

n^{2} + 3 n - 18 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 18) = 9

n_{1, 2} = \frac{- 3 \pm 9}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- 3 + 9}{2 \cdot 1}, n_{2} = \frac{- 3 - 9}{2 \cdot 1}

n = \frac{- 3 + 9}{2 \cdot 1} : 3

n = \frac{- 3 - 9}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

n = 3, n = - 6

- \frac{6}{9}

0 - y = 5

\frac{35 y ^{8} x ^{5}}{- 21 y ^{- 1} x ^{9}}

f (y) = y^{2} - 2

x - 3 = - 7

What are the solutions to the equation 6=n^2+3n-12 ?
The solutions to the equation 6=n^2+3n-12 are n=3,n=-6
Find the zeros of 6=n^2+3n-12
The zeros of 6=n^2+3n-12 are n=3,n=-6