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(429))/2 ,n=(sqrt(429)-19)/2

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

The zeros of 2n^2-19n+17=3n^2 are n=-(19+sqrt(429))/2 ,n=(sqrt(429)-19)/2

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

n = - \frac{19 + 429}{2}, n = \frac{429 - 19}{2}

Decimal Notation

n = - 19.85615 \dots, n = 0.85615 \dots

Solution steps

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

Move

3 n^{2}

to the left side

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

Solve with the quadratic formula

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

(- 19)^{2} - 4 (- 1) \cdot 17 = 429

n_{1, 2} = \frac{- ( - 19 ) \pm 429}{2 ( - 1 )}

Separate the solutions

n_{1} = \frac{- ( - 19 ) + 429}{2 ( - 1 )}, n_{2} = \frac{- ( - 19 ) - 429}{2 ( - 1 )}

n = \frac{- ( - 19 ) + 429}{2 ( - 1 )} : - \frac{19 + 429}{2}

n = \frac{- ( - 19 ) - 429}{2 ( - 1 )} : \frac{429 - 19}{2}

The solutions to the quadratic equation are:

n = - \frac{19 + 429}{2}, n = \frac{429 - 19}{2}

x^{2} + (x - 5)^{2} = 73

x^{2} = - 6, x^{1} = 1

2 y^{2} + y - 11 = (y + 1)^{2}

8 x^{2} - 2 x = 0

n^{2} = 3.6 \cdot 1 0^{13}

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(429))/2 ,n=(sqrt(429)-19)/2
Find the zeros of 2n^2-19n+17=3n^2
The zeros of 2n^2-19n+17=3n^2 are n=-(19+sqrt(429))/2 ,n=(sqrt(429)-19)/2