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

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

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

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

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

x = - 4, x = - 5

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 18 + 2}{2 \cdot 2} : - 4

x = \frac{- 18 - 2}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = - 4, x = - 5

5 (3 - x) \leq 3 x - 1

\frac{5}{8} x - 1 \leq 9

2 (5 c - 7) \geq 10 (c - 3)

6 x^{2} + 8 x - 8

9 x^{2} = 6 x + 3

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