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

The solutions to the equation r^2-2r-4=0 are r=1+sqrt(5),r=1-sqrt(5)

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

The zeros of r^2-2r-4=0 are r=1+sqrt(5),r=1-sqrt(5)

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r^{2} - 2 r - 4 = 0

r = 1 + 5, r = 1 - 5

Decimal Notation

r = 3.23606 \dots, r = - 1.23606 \dots

Solution steps

r^{2} - 2 r - 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 4) = 25

r_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 1}, r_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 1}

r = \frac{- ( - 2 ) + 2 5}{2 \cdot 1} : 1 + 5

r = \frac{- ( - 2 ) - 2 5}{2 \cdot 1} : 1 - 5

The solutions to the quadratic equation are:

r = 1 + 5, r = 1 - 5

\frac{1}{3} x - 5 + 171 = x

x^{2} - 6 x - 9 = - 6 x

\frac{x + 2}{3} = \frac{2 x - 4}{2}

(32)^{\frac{4}{5}}

What are the solutions to the equation r^2-2r-4=0 ?
The solutions to the equation r^2-2r-4=0 are r=1+sqrt(5),r=1-sqrt(5)
Find the zeros of r^2-2r-4=0
The zeros of r^2-2r-4=0 are r=1+sqrt(5),r=1-sqrt(5)