What are the solutions to the equation x^2+2x-288=0 ?

The solutions to the equation x^2+2x-288=0 are x=16,x=-18

Find the zeros of x^2+2x-288=0

The zeros of x^2+2x-288=0 are x=16,x=-18

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x^{2} + 2 x - 288 = 0

x = 16, x = - 18

Solution steps

x^{2} + 2 x - 288 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 288 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 288) = 34

x_{1, 2} = \frac{- 2 \pm 34}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 34}{2 \cdot 1}, x_{2} = \frac{- 2 - 34}{2 \cdot 1}

x = \frac{- 2 + 34}{2 \cdot 1} : 16

x = \frac{- 2 - 34}{2 \cdot 1} : - 18

The solutions to the quadratic equation are:

x = 16, x = - 18

94

\frac{x ^{- 2}}{y}

\frac{x}{6} - 2 = - 3

\frac{7 ^{8}}{7 ^{2}}

\frac{24}{3 2}

What are the solutions to the equation x^2+2x-288=0 ?
The solutions to the equation x^2+2x-288=0 are x=16,x=-18
Find the zeros of x^2+2x-288=0
The zeros of x^2+2x-288=0 are x=16,x=-18