What are the solutions to the equation 2r^2-3r-1=0 ?

The solutions to the equation 2r^2-3r-1=0 are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4

Find the zeros of 2r^2-3r-1=0

The zeros of 2r^2-3r-1=0 are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4

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2 r^{2} - 3 r - 1 = 0

r = \frac{3 + 17}{4}, r = \frac{3 - 17}{4}

Solution steps

2 r^{2} - 3 r - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

r = \frac{- ( - 3 ) + 17}{2 \cdot 2} : \frac{3 + 17}{4}

r = \frac{- ( - 3 ) - 17}{2 \cdot 2} : \frac{3 - 17}{4}

The solutions to the quadratic equation are:

r = \frac{3 + 17}{4}, r = \frac{3 - 17}{4}

3 x^{2} + 1 = 4 x

3 (x - 4)^{2} - 12

7 x + 14 = 9 x - 8

3 x^{3} + x^{2} - 38 x + 24

x^{2} + 2 x - 15 \leq 0

What are the solutions to the equation 2r^2-3r-1=0 ?
The solutions to the equation 2r^2-3r-1=0 are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4
Find the zeros of 2r^2-3r-1=0
The zeros of 2r^2-3r-1=0 are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4