What are the solutions to the equation 2n^2-19n+17=-3n^2 ?

The solutions to the equation 2n^2-19n+17=-3n^2 are n=(19+sqrt(21))/(10),n=(19-sqrt(21))/(10)

Find the zeros of 2n^2-19n+17=-3n^2

The zeros of 2n^2-19n+17=-3n^2 are n=(19+sqrt(21))/(10),n=(19-sqrt(21))/(10)

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2 n^{2} - 19 n + 17 = - 3 n^{2}

n = \frac{19 + 21}{10}, n = \frac{19 - 21}{10}

Solution steps

2 n^{2} - 19 n + 17 = - 3 n^{2}

Move

3 n^{2}

to the left side

5 n^{2} - 19 n + 17 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 5 \cdot 17}{2 \cdot 5}

(- 19)^{2} - 4 \cdot 5 \cdot 17 = 21

n_{1, 2} = \frac{- ( - 19 ) \pm 21}{2 \cdot 5}

Separate the solutions

n_{1} = \frac{- ( - 19 ) + 21}{2 \cdot 5}, n_{2} = \frac{- ( - 19 ) - 21}{2 \cdot 5}

n = \frac{- ( - 19 ) + 21}{2 \cdot 5} : \frac{19 + 21}{10}

n = \frac{- ( - 19 ) - 21}{2 \cdot 5} : \frac{19 - 21}{10}

The solutions to the quadratic equation are:

n = \frac{19 + 21}{10}, n = \frac{19 - 21}{10}

9 x^{2} + 18 x - 68 = 7

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

(5 m + 1)^{2} = 7

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

49 x^{2} + 9 = 42 x

What are the solutions to the equation 2n^2-19n+17=-3n^2 ?
The solutions to the equation 2n^2-19n+17=-3n^2 are n=(19+sqrt(21))/(10),n=(19-sqrt(21))/(10)
Find the zeros of 2n^2-19n+17=-3n^2
The zeros of 2n^2-19n+17=-3n^2 are n=(19+sqrt(21))/(10),n=(19-sqrt(21))/(10)