What are the solutions to the equation x^2+2x-7=17 ?

The solutions to the equation x^2+2x-7=17 are x=4,x=-6

Find the zeros of x^2+2x-7=17

The zeros of x^2+2x-7=17 are x=4,x=-6

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x^{2} + 2 x - 7 = 17

x = 4, x = - 6

Solution steps

x^{2} + 2 x - 7 = 17

Move

17

to the left side

x^{2} + 2 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 24) = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 \cdot 1}, x_{2} = \frac{- 2 - 10}{2 \cdot 1}

x = \frac{- 2 + 10}{2 \cdot 1} : 4

x = \frac{- 2 - 10}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

x = 4, x = - 6

- 3 a^{2} - 8 a - 5 = 0

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

n^{2} - 4 n = 11

x^{2} - 7 x - 34 = 0

x^{2} + 289 = 0

What are the solutions to the equation x^2+2x-7=17 ?
The solutions to the equation x^2+2x-7=17 are x=4,x=-6
Find the zeros of x^2+2x-7=17
The zeros of x^2+2x-7=17 are x=4,x=-6