What are the solutions to the equation 2x^2+28x-480=0 ?

The solutions to the equation 2x^2+28x-480=0 are x=10,x=-24

Find the zeros of 2x^2+28x-480=0

The zeros of 2x^2+28x-480=0 are x=10,x=-24

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2 x^{2} + 28 x - 480 = 0

x = 10, x = - 24

Solution steps

2 x^{2} + 28 x - 480 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 2 ( - 480 )}{2 \cdot 2}

2 8^{2} - 4 \cdot 2 (- 480) = 68

x_{1, 2} = \frac{- 28 \pm 68}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 28 + 68}{2 \cdot 2}, x_{2} = \frac{- 28 - 68}{2 \cdot 2}

x = \frac{- 28 + 68}{2 \cdot 2} : 10

x = \frac{- 28 - 68}{2 \cdot 2} : - 24

The solutions to the quadratic equation are:

x = 10, x = - 24

a^{2} + 14 a + 63 = 8

\frac{x ^{2} - 4 x}{x + 5} > 0

(- 4 a^{2}) (2 a^{- 3})^{- 4}

8 + x > 31

2 \cdot 5 - 2 \cdot 5 + 2 \cdot 2 \cdot 5 + 10

What are the solutions to the equation 2x^2+28x-480=0 ?
The solutions to the equation 2x^2+28x-480=0 are x=10,x=-24
Find the zeros of 2x^2+28x-480=0
The zeros of 2x^2+28x-480=0 are x=10,x=-24