What are the solutions to the equation 5r^2-12r+2=-3r ?

The solutions to the equation 5r^2-12r+2=-3r are r=(9+sqrt(41))/(10),r=(9-sqrt(41))/(10)

Find the zeros of 5r^2-12r+2=-3r

The zeros of 5r^2-12r+2=-3r are r=(9+sqrt(41))/(10),r=(9-sqrt(41))/(10)

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5 r^{2} - 12 r + 2 = - 3 r

r = \frac{9 + 41}{10}, r = \frac{9 - 41}{10}

Decimal Notation

r = 1.54031 \dots, r = 0.25968 \dots

Solution steps

5 r^{2} - 12 r + 2 = - 3 r

Move

3 r

to the left side

5 r^{2} - 9 r + 2 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 \cdot 2}{2 \cdot 5}

(- 9)^{2} - 4 \cdot 5 \cdot 2 = 41

r_{1, 2} = \frac{- ( - 9 ) \pm 41}{2 \cdot 5}

Separate the solutions

r_{1} = \frac{- ( - 9 ) + 41}{2 \cdot 5}, r_{2} = \frac{- ( - 9 ) - 41}{2 \cdot 5}

r = \frac{- ( - 9 ) + 41}{2 \cdot 5} : \frac{9 + 41}{10}

r = \frac{- ( - 9 ) - 41}{2 \cdot 5} : \frac{9 - 41}{10}

The solutions to the quadratic equation are:

r = \frac{9 + 41}{10}, r = \frac{9 - 41}{10}

2 p^{2} - 6 p - 56 = 0

(4 y + 12)^{2} = 32

6 x^{2} - 6 x + 9 = 0

x^{2} - 2 x + 1 - 2 = 0

- x^{2} = - 25

What are the solutions to the equation 5r^2-12r+2=-3r ?
The solutions to the equation 5r^2-12r+2=-3r are r=(9+sqrt(41))/(10),r=(9-sqrt(41))/(10)
Find the zeros of 5r^2-12r+2=-3r
The zeros of 5r^2-12r+2=-3r are r=(9+sqrt(41))/(10),r=(9-sqrt(41))/(10)