What are the solutions to the equation 4x^2+72x+320=0 ?

The solutions to the equation 4x^2+72x+320=0 are x=-8,x=-10

Find the zeros of 4x^2+72x+320=0

The zeros of 4x^2+72x+320=0 are x=-8,x=-10

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4 x^{2} + 72 x + 320 = 0

x = - 8, x = - 10

Solution steps

4 x^{2} + 72 x + 320 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 72 \pm 7 2 ^{2} - 4 \cdot 4 \cdot 320}{2 \cdot 4}

7 2^{2} - 4 \cdot 4 \cdot 320 = 8

x_{1, 2} = \frac{- 72 \pm 8}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 72 + 8}{2 \cdot 4}, x_{2} = \frac{- 72 - 8}{2 \cdot 4}

x = \frac{- 72 + 8}{2 \cdot 4} : - 8

x = \frac{- 72 - 8}{2 \cdot 4} : - 10

The solutions to the quadratic equation are:

x = - 8, x = - 10

1 \cdot 1 0^{- 4} = \frac{1}{2} \cdot C \cdot (8)^{2}

- 2 x > - 8

2 x^{2} - 3 x + 6 = 0

\frac{x}{7} = 3.3

\frac{- 18}{- 3}

What are the solutions to the equation 4x^2+72x+320=0 ?
The solutions to the equation 4x^2+72x+320=0 are x=-8,x=-10
Find the zeros of 4x^2+72x+320=0
The zeros of 4x^2+72x+320=0 are x=-8,x=-10