What are the solutions to the equation x^2+x-240=0 ?

The solutions to the equation x^2+x-240=0 are x=15,x=-16

Find the zeros of x^2+x-240=0

The zeros of x^2+x-240=0 are x=15,x=-16

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x^{2} + x - 240 = 0

x = 15, x = - 16

Solution steps

x^{2} + x - 240 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 240 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 240) = 31

x_{1, 2} = \frac{- 1 \pm 31}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 31}{2 \cdot 1}, x_{2} = \frac{- 1 - 31}{2 \cdot 1}

x = \frac{- 1 + 31}{2 \cdot 1} : 15

x = \frac{- 1 - 31}{2 \cdot 1} : - 16

The solutions to the quadratic equation are:

x = 15, x = - 16

- 6 + 3

(x^{2})^{2}

8 - \frac{9 x}{2} \leq 5

- (8)^{2}

- \frac{1}{3} - \frac{1}{2} + 2

What are the solutions to the equation x^2+x-240=0 ?
The solutions to the equation x^2+x-240=0 are x=15,x=-16
Find the zeros of x^2+x-240=0
The zeros of x^2+x-240=0 are x=15,x=-16