What are the solutions to the equation x^2+2x-1680=0 ?

The solutions to the equation x^2+2x-1680=0 are x=40,x=-42

Find the zeros of x^2+2x-1680=0

The zeros of x^2+2x-1680=0 are x=40,x=-42

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x^{2} + 2 x - 1680 = 0

x = 40, x = - 42

Solution steps

x^{2} + 2 x - 1680 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 1680 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 1680) = 82

x_{1, 2} = \frac{- 2 \pm 82}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 82}{2 \cdot 1}, x_{2} = \frac{- 2 - 82}{2 \cdot 1}

x = \frac{- 2 + 82}{2 \cdot 1} : 40

x = \frac{- 2 - 82}{2 \cdot 1} : - 42

The solutions to the quadratic equation are:

x = 40, x = - 42

3 x^{2} - 6 x - 15 = 12 x + 12

x, x^{2} y + 4 y - 8 = 0

(x - 8)^{2} = - 81

12 x^{2} - 88 x + 112 = 0

x, 25 x^{2} + 9 y^{2} + 18 y + 216 = 0

What are the solutions to the equation x^2+2x-1680=0 ?
The solutions to the equation x^2+2x-1680=0 are x=40,x=-42
Find the zeros of x^2+2x-1680=0
The zeros of x^2+2x-1680=0 are x=40,x=-42