What are the solutions to the equation 2r^2-6r-1=-3r ?

The solutions to the equation 2r^2-6r-1=-3r are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4

Find the zeros of 2r^2-6r-1=-3r

The zeros of 2r^2-6r-1=-3r are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4

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2 r^{2} - 6 r - 1 = - 3 r

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

Decimal Notation

r = 1.78077 \dots, r = - 0.28077 \dots

Solution steps

2 r^{2} - 6 r - 1 = - 3 r

Move

3 r

to the left side

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}

5 x^{2} + 13 x + 30 = 0

7 x^{2} + 56 x = 0

2 b^{2} + 5 b - 2 = 6 b

4 x^{2} - 4 + 5 x = 0

3 x^{2} = 39 x

What are the solutions to the equation 2r^2-6r-1=-3r ?
The solutions to the equation 2r^2-6r-1=-3r are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4
Find the zeros of 2r^2-6r-1=-3r
The zeros of 2r^2-6r-1=-3r are r=(3+sqrt(17))/4 ,r=(3-sqrt(17))/4