What are the solutions to the equation n(n-3)=40 ?

The solutions to the equation n(n-3)=40 are n=8,n=-5

Find the zeros of n(n-3)=40

The zeros of n(n-3)=40 are n=8,n=-5

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n (n - 3) = 40

n = 8, n = - 5

Solution steps

n (n - 3) = 40

Expand

n (n - 3) : n^{2} - 3 n

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

y, y^{2} + a^{2} = (a + y) 2 - a (a + 1)

a = (2 x + 1) 2 - (2 x - 1) 2 - (2 - x) (2 + x) - (x + 1) 2 + 5

x^{2} + x + 100.25 = 0

27 = 2 x - \frac{1}{73} x^{2}

A (x) = (x + 1)^{2} - 36

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