What are the solutions to the equation 2n^2-5n=-290 ?

The solutions to the equation 2n^2-5n=-290 are n= 5/4+i(3sqrt(255))/4 ,n= 5/4-i(3sqrt(255))/4

Find the zeros of 2n^2-5n=-290

The zeros of 2n^2-5n=-290 are n= 5/4+i(3sqrt(255))/4 ,n= 5/4-i(3sqrt(255))/4

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2 n^{2} - 5 n = - 290

n = \frac{5}{4} + i \frac{3 255}{4}, n = \frac{5}{4} - i \frac{3 255}{4}

Solution steps

2 n^{2} - 5 n = - 290

Move

290

to the left side

2 n^{2} - 5 n + 290 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 \cdot 290}{2 \cdot 2}

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 290 : 3255 i

n_{1, 2} = \frac{- ( - 5 ) \pm 3 255 i}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 5 ) + 3 255 i}{2 \cdot 2}, n_{2} = \frac{- ( - 5 ) - 3 255 i}{2 \cdot 2}

n = \frac{- ( - 5 ) + 3 255 i}{2 \cdot 2} : \frac{5}{4} + i \frac{3 255}{4}

n = \frac{- ( - 5 ) - 3 255 i}{2 \cdot 2} : \frac{5}{4} - i \frac{3 255}{4}

The solutions to the quadratic equation are:

n = \frac{5}{4} + i \frac{3 255}{4}, n = \frac{5}{4} - i \frac{3 255}{4}

5 x^{2} = 140 + 15 x

54 = 6 (2 x - 1)^{2}

7 x^{2} - 28 x + 28 = 0

6 x^{2} + 5 = 17 x

5 x + 6 - x^{2} = x^{2} - 2 x - 3

What are the solutions to the equation 2n^2-5n=-290 ?
The solutions to the equation 2n^2-5n=-290 are n= 5/4+i(3sqrt(255))/4 ,n= 5/4-i(3sqrt(255))/4
Find the zeros of 2n^2-5n=-290
The zeros of 2n^2-5n=-290 are n= 5/4+i(3sqrt(255))/4 ,n= 5/4-i(3sqrt(255))/4