What are the solutions to the equation n^2-3n=40 ?

The solutions to the equation n^2-3n=40 are n=8,n=-5

Find the zeros of n^2-3n=40

The zeros of n^2-3n=40 are n=8,n=-5

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n^{2} - 3 n = 40

n = 8, n = - 5

Solution steps

n^{2} - 3 n = 40

Move

40

to the left side

n^{2} - 3 n - 40 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 40 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 40) = 13

n_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 1}, n_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 1}

n = \frac{- ( - 3 ) + 13}{2 \cdot 1} : 8

n = \frac{- ( - 3 ) - 13}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

n = 8, n = - 5

- 3 x^{2} + 75 x = 0

4 y^{2} + 12 y + 25 = 3 y^{2} - 10

x, 15 = 2 x^{2} - 3 x + 1

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

(5 x - 3)^{2} = - 3

What are the solutions to the equation n^2-3n=40 ?
The solutions to the equation n^2-3n=40 are n=8,n=-5
Find the zeros of n^2-3n=40
The zeros of n^2-3n=40 are n=8,n=-5