What are the solutions to the equation r^2-3r-2=0 ?

The solutions to the equation r^2-3r-2=0 are r=(3+sqrt(17))/2 ,r=(3-sqrt(17))/2

Find the zeros of r^2-3r-2=0

The zeros of r^2-3r-2=0 are r=(3+sqrt(17))/2 ,r=(3-sqrt(17))/2

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r^{2} - 3 r - 2 = 0

r = \frac{3 + 17}{2}, r = \frac{3 - 17}{2}

Decimal Notation

r = 3.56155 \dots, r = - 0.56155 \dots

Solution steps

r^{2} - 3 r - 2 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 2 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 2) = 17

r_{1, 2} = \frac{- ( - 3 ) \pm 17}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 17}{2 \cdot 1}, r_{2} = \frac{- ( - 3 ) - 17}{2 \cdot 1}

r = \frac{- ( - 3 ) + 17}{2 \cdot 1} : \frac{3 + 17}{2}

r = \frac{- ( - 3 ) - 17}{2 \cdot 1} : \frac{3 - 17}{2}

The solutions to the quadratic equation are:

r = \frac{3 + 17}{2}, r = \frac{3 - 17}{2}

- 2 x^{2} + 2 x + 6 = 0

- 2 x^{2} + 10 x + 48 = 0

2 x^{2} - x - 6 = 0

5 x^{2} + 2 = 6 x^{2} - 4

6 u^{2} - 6 = 0

What are the solutions to the equation r^2-3r-2=0 ?
The solutions to the equation r^2-3r-2=0 are r=(3+sqrt(17))/2 ,r=(3-sqrt(17))/2
Find the zeros of r^2-3r-2=0
The zeros of r^2-3r-2=0 are r=(3+sqrt(17))/2 ,r=(3-sqrt(17))/2