What are the solutions to the equation 12n^2-6n-24=0 ?

The solutions to the equation 12n^2-6n-24=0 are n=(1+sqrt(33))/4 ,n=(1-sqrt(33))/4

Find the zeros of 12n^2-6n-24=0

The zeros of 12n^2-6n-24=0 are n=(1+sqrt(33))/4 ,n=(1-sqrt(33))/4

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12 n^{2} - 6 n - 24 = 0

n = \frac{1 + 33}{4}, n = \frac{1 - 33}{4}

Decimal Notation

n = 1.68614 \dots, n = - 1.18614 \dots

Solution steps

12 n^{2} - 6 n - 24 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 12 ( - 24 )}{2 \cdot 12}

(- 6)^{2} - 4 \cdot 12 (- 24) = 633

n_{1, 2} = \frac{- ( - 6 ) \pm 6 33}{2 \cdot 12}

Separate the solutions

n_{1} = \frac{- ( - 6 ) + 6 33}{2 \cdot 12}, n_{2} = \frac{- ( - 6 ) - 6 33}{2 \cdot 12}

n = \frac{- ( - 6 ) + 6 33}{2 \cdot 12} : \frac{1 + 33}{4}

n = \frac{- ( - 6 ) - 6 33}{2 \cdot 12} : \frac{1 - 33}{4}

The solutions to the quadratic equation are:

n = \frac{1 + 33}{4}, n = \frac{1 - 33}{4}

\frac{i ^{26} - i}{i - 1}

3 x + 6 = 6 x

x^{2} + 14 x + 22 = 0

- \frac{1.1 ( 5 )}{( 0.01 ( 5 ) ^{2} + 1 ) ^{2}}

3 (x - 1) + 7 > 4 x + 2 - 3 (2 - x)

What are the solutions to the equation 12n^2-6n-24=0 ?
The solutions to the equation 12n^2-6n-24=0 are n=(1+sqrt(33))/4 ,n=(1-sqrt(33))/4
Find the zeros of 12n^2-6n-24=0
The zeros of 12n^2-6n-24=0 are n=(1+sqrt(33))/4 ,n=(1-sqrt(33))/4