What are the solutions to the equation 2x^2+28x-64=0 ?

The solutions to the equation 2x^2+28x-64=0 are x=2,x=-16

Find the zeros of 2x^2+28x-64=0

The zeros of 2x^2+28x-64=0 are x=2,x=-16

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2 x^{2} + 28 x - 64 = 0

x = 2, x = - 16

Solution steps

2 x^{2} + 28 x - 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 2 ( - 64 )}{2 \cdot 2}

2 8^{2} - 4 \cdot 2 (- 64) = 36

x_{1, 2} = \frac{- 28 \pm 36}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 28 + 36}{2 \cdot 2}, x_{2} = \frac{- 28 - 36}{2 \cdot 2}

x = \frac{- 28 + 36}{2 \cdot 2} : 2

x = \frac{- 28 - 36}{2 \cdot 2} : - 16

The solutions to the quadratic equation are:

x = 2, x = - 16

- 6 = - 5 x^{2} - 13 x

x^{2} + 10 x + 25 = 0

5 x^{2} + 6 x - 12 = 0

x^{2} - 6 x + 9 x = 0

\frac{- 1543}{120} = \frac{3}{5} x \frac{11}{4} (- \frac{11}{5} x \frac{8}{5})

What are the solutions to the equation 2x^2+28x-64=0 ?
The solutions to the equation 2x^2+28x-64=0 are x=2,x=-16
Find the zeros of 2x^2+28x-64=0
The zeros of 2x^2+28x-64=0 are x=2,x=-16