What are the solutions to the equation 2x^2+20x+48=0 ?

The solutions to the equation 2x^2+20x+48=0 are x=-4,x=-6

Find the zeros of 2x^2+20x+48=0

The zeros of 2x^2+20x+48=0 are x=-4,x=-6

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2 x^{2} + 20 x + 48 = 0

x = - 4, x = - 6

Solution steps

2 x^{2} + 20 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 2 \cdot 48}{2 \cdot 2}

2 0^{2} - 4 \cdot 2 \cdot 48 = 4

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 4, x = - 6

x^{2} - 10 = - 18

18 x^{2} - 72 = 0

x^{2} - 3 + 2 = 0

- x^{2} + 2 x - 7 = 0

x^{2} = 2 x^{2}

What are the solutions to the equation 2x^2+20x+48=0 ?
The solutions to the equation 2x^2+20x+48=0 are x=-4,x=-6
Find the zeros of 2x^2+20x+48=0
The zeros of 2x^2+20x+48=0 are x=-4,x=-6