What are the solutions to the equation 2n^2-10n-5=-4n ?

The solutions to the equation 2n^2-10n-5=-4n are n=(3+sqrt(19))/2 ,n=(3-sqrt(19))/2

Find the zeros of 2n^2-10n-5=-4n

The zeros of 2n^2-10n-5=-4n are n=(3+sqrt(19))/2 ,n=(3-sqrt(19))/2

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2 n^{2} - 10 n - 5 = - 4 n

n = \frac{3 + 19}{2}, n = \frac{3 - 19}{2}

Decimal Notation

n = 3.67944 \dots, n = - 0.67944 \dots

Solution steps

2 n^{2} - 10 n - 5 = - 4 n

Move

4 n

to the left side

2 n^{2} - 6 n - 5 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 6)^{2} - 4 \cdot 2 (- 5) = 219

n_{1, 2} = \frac{- ( - 6 ) \pm 2 19}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 6 ) + 2 19}{2 \cdot 2}, n_{2} = \frac{- ( - 6 ) - 2 19}{2 \cdot 2}

n = \frac{- ( - 6 ) + 2 19}{2 \cdot 2} : \frac{3 + 19}{2}

n = \frac{- ( - 6 ) - 2 19}{2 \cdot 2} : \frac{3 - 19}{2}

The solutions to the quadratic equation are:

n = \frac{3 + 19}{2}, n = \frac{3 - 19}{2}

2 x^{2} - x = 45

6 - x - x^{2} = 0

(x + 6) (x + 4) = 3

t^{2} - 45 = 0

- 8 b^{2} - 3 b + 22 = 0

What are the solutions to the equation 2n^2-10n-5=-4n ?
The solutions to the equation 2n^2-10n-5=-4n are n=(3+sqrt(19))/2 ,n=(3-sqrt(19))/2
Find the zeros of 2n^2-10n-5=-4n
The zeros of 2n^2-10n-5=-4n are n=(3+sqrt(19))/2 ,n=(3-sqrt(19))/2