What are the solutions to the equation 2n^2-18n+40=0 ?

The solutions to the equation 2n^2-18n+40=0 are n=5,n=4

Find the zeros of 2n^2-18n+40=0

The zeros of 2n^2-18n+40=0 are n=5,n=4

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2 n^{2} - 18 n + 40 = 0

n = 5, n = 4

Solution steps

2 n^{2} - 18 n + 40 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 2 \cdot 40}{2 \cdot 2}

(- 18)^{2} - 4 \cdot 2 \cdot 40 = 2

n_{1, 2} = \frac{- ( - 18 ) \pm 2}{2 \cdot 2}

Separate the solutions

n_{1} = \frac{- ( - 18 ) + 2}{2 \cdot 2}, n_{2} = \frac{- ( - 18 ) - 2}{2 \cdot 2}

n = \frac{- ( - 18 ) + 2}{2 \cdot 2} : 5

n = \frac{- ( - 18 ) - 2}{2 \cdot 2} : 4

The solutions to the quadratic equation are:

n = 5, n = 4

729 = 9 x^{2}

x^{2} - 94 = - x - 4

6 y^{2} - 13 y + 12 = 4 y^{2}

2 x^{2} - 13 = - 5 x - 10

- x^{2} + 6 x - 14 = 0

What are the solutions to the equation 2n^2-18n+40=0 ?
The solutions to the equation 2n^2-18n+40=0 are n=5,n=4
Find the zeros of 2n^2-18n+40=0
The zeros of 2n^2-18n+40=0 are n=5,n=4