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

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

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

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

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40 = n^{2} - 3 n

n = 8, n = - 5

Solution steps

40 = n^{2} - 3 n

Switch sides

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

x, y^{3} + y^{2} + 5 x y + x^{2} + x + y = 0

3 t (t + 1) = 4 (t + 1)

x^{2} + 22 x + 20 = 3 x^{2}

x^{2} + 12 x + 63 = 0

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

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