What are the solutions to the equation 2x^2+4x=96 ?

The solutions to the equation 2x^2+4x=96 are x=6,x=-8

Find the zeros of 2x^2+4x=96

The zeros of 2x^2+4x=96 are x=6,x=-8

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2 x^{2} + 4 x = 96

x = 6, x = - 8

Solution steps

2 x^{2} + 4 x = 96

Move

96

to the left side

2 x^{2} + 4 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 96 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 96) = 28

x_{1, 2} = \frac{- 4 \pm 28}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 28}{2 \cdot 2}, x_{2} = \frac{- 4 - 28}{2 \cdot 2}

x = \frac{- 4 + 28}{2 \cdot 2} : 6

x = \frac{- 4 - 28}{2 \cdot 2} : - 8

The solutions to the quadratic equation are:

x = 6, x = - 8

x - x^{2} = 6

\frac{4}{\frac{1}{2}}

\frac{4}{5} (10 u + 65) \geq \frac{6}{5} (5 u - 10) + 4 u

(- x)^{2}

\frac{7 x}{2} - 3 = \frac{8 x}{3} + 5

What are the solutions to the equation 2x^2+4x=96 ?
The solutions to the equation 2x^2+4x=96 are x=6,x=-8
Find the zeros of 2x^2+4x=96
The zeros of 2x^2+4x=96 are x=6,x=-8