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

The solutions to the equation a^2-a-4=0 are a=(1+sqrt(17))/2 ,a=(1-sqrt(17))/2

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

The zeros of a^2-a-4=0 are a=(1+sqrt(17))/2 ,a=(1-sqrt(17))/2

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a^{2} - a - 4 = 0

a = \frac{1 + 17}{2}, a = \frac{1 - 17}{2}

Decimal Notation

a = 2.56155 \dots, a = - 1.56155 \dots

Solution steps

a^{2} - a - 4 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 4) = 17

a_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 1}, a_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 1}

a = \frac{- ( - 1 ) + 17}{2 \cdot 1} : \frac{1 + 17}{2}

a = \frac{- ( - 1 ) - 17}{2 \cdot 1} : \frac{1 - 17}{2}

The solutions to the quadratic equation are:

a = \frac{1 + 17}{2}, a = \frac{1 - 17}{2}

x^{2} - 12 x + 36 = 64

c, (c + y) (c - y) = (3 c^{2} - y) y

1 8^{2} + b^{2} = 3 0^{2}

x (x + 3) = 270

a^{2} + 2 a + 3 = 0

What are the solutions to the equation a^2-a-4=0 ?
The solutions to the equation a^2-a-4=0 are a=(1+sqrt(17))/2 ,a=(1-sqrt(17))/2
Find the zeros of a^2-a-4=0
The zeros of a^2-a-4=0 are a=(1+sqrt(17))/2 ,a=(1-sqrt(17))/2