What are the solutions to the equation g^2+4g-32=0 ?

The solutions to the equation g^2+4g-32=0 are g=4,g=-8

Find the zeros of g^2+4g-32=0

The zeros of g^2+4g-32=0 are g=4,g=-8

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g^{2} + 4 g - 32 = 0

g = 4, g = - 8

Solution steps

g^{2} + 4 g - 32 = 0

Solve with the quadratic formula

g_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 32) = 12

g_{1, 2} = \frac{- 4 \pm 12}{2 \cdot 1}

Separate the solutions

g_{1} = \frac{- 4 + 12}{2 \cdot 1}, g_{2} = \frac{- 4 - 12}{2 \cdot 1}

g = \frac{- 4 + 12}{2 \cdot 1} : 4

g = \frac{- 4 - 12}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

g = 4, g = - 8

x, x^{2} + 2 x = 3

x^{2} - 25 x + 4 = 0

- x^{2} + 120 x - 2975 = 0

5 n^{2} + 2 n - 8 = 0

(x^{2} - 3 x + 9) = 0

What are the solutions to the equation g^2+4g-32=0 ?
The solutions to the equation g^2+4g-32=0 are g=4,g=-8
Find the zeros of g^2+4g-32=0
The zeros of g^2+4g-32=0 are g=4,g=-8