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

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

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

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

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

a = 8, a = - 4

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

a = 8, a = - 4

(4 x - 3)^{2} - (x - 1) (x - 9) = 0

x^{2} + 10 x + 28

3 z^{2} - 8 z + 4 = 5 z

\frac{y ^{2}}{4} = \frac{3 y + 4}{4}

x^{2} + 8 x - 11 = 6 x - 8

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