What are the solutions to the equation 4n^2-20=-12n ?

The solutions to the equation 4n^2-20=-12n are n=(-3+sqrt(29))/2 ,n=-(3+sqrt(29))/2

Find the zeros of 4n^2-20=-12n

The zeros of 4n^2-20=-12n are n=(-3+sqrt(29))/2 ,n=-(3+sqrt(29))/2

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4 n^{2} - 20 = - 12 n

n = \frac{- 3 + 29}{2}, n = - \frac{3 + 29}{2}

Decimal Notation

n = 1.19258 \dots, n = - 4.19258 \dots

Solution steps

4 n^{2} - 20 = - 12 n

Move

12 n

to the left side

4 n^{2} - 20 + 12 n = 0

Write in the standard form

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

4 n^{2} + 12 n - 20 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 ( - 20 )}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 (- 20) = 429

n_{1, 2} = \frac{- 12 \pm 4 29}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- 12 + 4 29}{2 \cdot 4}, n_{2} = \frac{- 12 - 4 29}{2 \cdot 4}

n = \frac{- 12 + 4 29}{2 \cdot 4} : \frac{- 3 + 29}{2}

n = \frac{- 12 - 4 29}{2 \cdot 4} : - \frac{3 + 29}{2}

The solutions to the quadratic equation are:

n = \frac{- 3 + 29}{2}, n = - \frac{3 + 29}{2}

2 x^{2} - 2 x - 16 = 0

x^{2} - 4 x + 36 = 0

3 x^{2} - 4 x + 5 = 4

2 x^{2} + 5 x - 11 = 1

s^{2} + 14 s + 2

What are the solutions to the equation 4n^2-20=-12n ?
The solutions to the equation 4n^2-20=-12n are n=(-3+sqrt(29))/2 ,n=-(3+sqrt(29))/2
Find the zeros of 4n^2-20=-12n
The zeros of 4n^2-20=-12n are n=(-3+sqrt(29))/2 ,n=-(3+sqrt(29))/2