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

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

Find the zeros of h^2-4h-32=0

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

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h^{2} - 4 h - 32 = 0

h = 8, h = - 4

Solution steps

h^{2} - 4 h - 32 = 0

Solve with the quadratic formula

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

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

h_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 1}

Separate the solutions

h_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 1}, h_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 1}

h = \frac{- ( - 4 ) + 12}{2 \cdot 1} : 8

h = \frac{- ( - 4 ) - 12}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

h = 8, h = - 4

G (x) = x^{2} + 2

y^{2} - 4 y + 9 = 0

y^{2} - 144 = 0

14 y^{2} - 3 = - 19 y

x^{2} = 2 x

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