What are the solutions to the equation 5x^2+40x+80=0 ?

The solutions to the equation 5x^2+40x+80=0 are x=-4

Find the zeros of 5x^2+40x+80=0

The zeros of 5x^2+40x+80=0 are x=-4

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5 x^{2} + 40 x + 80 = 0

x = - 4

Solution steps

5 x^{2} + 40 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 5 \cdot 80}{2 \cdot 5}

4 0^{2} - 4 \cdot 5 \cdot 80 = 0

x_{1, 2} = \frac{- 40 \pm 0}{2 \cdot 5}

x = \frac{- 40}{2 \cdot 5}

\frac{- 40}{2 \cdot 5} = - 4

x = - 4

The solution to the quadratic equation is:

x = - 4

2 x - \frac{5}{3} = 7

- \frac{ln ( \frac{4}{5} )}{ln ( \frac{1.05}{1.035} )}

\frac{w ^{2}}{2} - w = \frac{3}{4}

b^{2} - 8 b - 20 = 0

What are the solutions to the equation 5x^2+40x+80=0 ?
The solutions to the equation 5x^2+40x+80=0 are x=-4
Find the zeros of 5x^2+40x+80=0
The zeros of 5x^2+40x+80=0 are x=-4