What are the solutions to the equation 2x^2+8x-42=0 ?

The solutions to the equation 2x^2+8x-42=0 are x=3,x=-7

Find the zeros of 2x^2+8x-42=0

The zeros of 2x^2+8x-42=0 are x=3,x=-7

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2 x^{2} + 8 x - 42 = 0

x = 3, x = - 7

Solution steps

2 x^{2} + 8 x - 42 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 42 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 42) = 20

x_{1, 2} = \frac{- 8 \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 8 + 20}{2 \cdot 2}, x_{2} = \frac{- 8 - 20}{2 \cdot 2}

x = \frac{- 8 + 20}{2 \cdot 2} : 3

x = \frac{- 8 - 20}{2 \cdot 2} : - 7

The solutions to the quadratic equation are:

x = 3, x = - 7

3 x^{2} - 8 x - 20 = 0

x^{2} - 9 x + 18 = 0

4 n^{2} - 10 n + 3 = - 2

5 x^{2} - 60 x + 360 = 0

1 + x + x^{2} = 0

What are the solutions to the equation 2x^2+8x-42=0 ?
The solutions to the equation 2x^2+8x-42=0 are x=3,x=-7
Find the zeros of 2x^2+8x-42=0
The zeros of 2x^2+8x-42=0 are x=3,x=-7