What are the solutions to the equation 2x^2+4x=48 ?

The solutions to the equation 2x^2+4x=48 are x=4,x=-6

Find the zeros of 2x^2+4x=48

The zeros of 2x^2+4x=48 are x=4,x=-6

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2 x^{2} + 4 x = 48

x = 4, x = - 6

Solution steps

2 x^{2} + 4 x = 48

Move

48

to the left side

2 x^{2} + 4 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 48 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 48) = 20

x_{1, 2} = \frac{- 4 \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 20}{2 \cdot 2}, x_{2} = \frac{- 4 - 20}{2 \cdot 2}

x = \frac{- 4 + 20}{2 \cdot 2} : 4

x = \frac{- 4 - 20}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

x = 4, x = - 6

- 10 x - 5 = - 8 x + 11

2 X - 7 = 3

2 x y - 3 x z

\frac{6}{4 3}

x^{3} + 8 x^{2} + 11 x - 20

What are the solutions to the equation 2x^2+4x=48 ?
The solutions to the equation 2x^2+4x=48 are x=4,x=-6
Find the zeros of 2x^2+4x=48
The zeros of 2x^2+4x=48 are x=4,x=-6