What are the solutions to the equation x^2+4x-320=0 ?

The solutions to the equation x^2+4x-320=0 are x=16,x=-20

Find the zeros of x^2+4x-320=0

The zeros of x^2+4x-320=0 are x=16,x=-20

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x^{2} + 4 x - 320 = 0

x = 16, x = - 20

Solution steps

x^{2} + 4 x - 320 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 320 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 320) = 36

x_{1, 2} = \frac{- 4 \pm 36}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 36}{2 \cdot 1}, x_{2} = \frac{- 4 - 36}{2 \cdot 1}

x = \frac{- 4 + 36}{2 \cdot 1} : 16

x = \frac{- 4 - 36}{2 \cdot 1} : - 20

The solutions to the quadratic equation are:

x = 16, x = - 20

(2 r + 1)^{2} + (r - 3)^{2} = 6

12 x^{2} - 8 x - 4 = 0

7 x^{2} + 56 x - 22 = 0

x^{2} = 5

q^{2} = - 4

What are the solutions to the equation x^2+4x-320=0 ?
The solutions to the equation x^2+4x-320=0 are x=16,x=-20
Find the zeros of x^2+4x-320=0
The zeros of x^2+4x-320=0 are x=16,x=-20