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

The solutions to the equation x^2+18x-40=0 are x=2,x=-20

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

The zeros of x^2+18x-40=0 are x=2,x=-20

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

x = 2, x = - 20

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 1 \cdot ( - 40 )}{2 \cdot 1}

1 8^{2} - 4 \cdot 1 \cdot (- 40) = 22

x_{1, 2} = \frac{- 18 \pm 22}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 18 + 22}{2 \cdot 1}, x_{2} = \frac{- 18 - 22}{2 \cdot 1}

x = \frac{- 18 + 22}{2 \cdot 1} : 2

x = \frac{- 18 - 22}{2 \cdot 1} : - 20

The solutions to the quadratic equation are:

x = 2, x = - 20

x (x + 4) = - 29

4 x^{2} - 4 x + 1

2 (x - 1)^{2} = 5

n^{2} - 10 n + 21 = 0

x^{2} + 12 x + 4 = 0

What are the solutions to the equation x^2+18x-40=0 ?
The solutions to the equation x^2+18x-40=0 are x=2,x=-20
Find the zeros of x^2+18x-40=0
The zeros of x^2+18x-40=0 are x=2,x=-20