What are the solutions to the equation 2r^2+7r-1=-3r^2 ?

The solutions to the equation 2r^2+7r-1=-3r^2 are r=(-7+sqrt(69))/(10),r=(-7-sqrt(69))/(10)

Find the zeros of 2r^2+7r-1=-3r^2

The zeros of 2r^2+7r-1=-3r^2 are r=(-7+sqrt(69))/(10),r=(-7-sqrt(69))/(10)

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2 r^{2} + 7 r - 1 = - 3 r^{2}

r = \frac{- 7 + 69}{10}, r = \frac{- 7 - 69}{10}

Solution steps

2 r^{2} + 7 r - 1 = - 3 r^{2}

Move

3 r^{2}

to the left side

5 r^{2} + 7 r - 1 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 5 ( - 1 )}{2 \cdot 5}

7^{2} - 4 \cdot 5 (- 1) = 69

r_{1, 2} = \frac{- 7 \pm 69}{2 \cdot 5}

Separate the solutions

r_{1} = \frac{- 7 + 69}{2 \cdot 5}, r_{2} = \frac{- 7 - 69}{2 \cdot 5}

r = \frac{- 7 + 69}{2 \cdot 5} : \frac{- 7 + 69}{10}

r = \frac{- 7 - 69}{2 \cdot 5} : \frac{- 7 - 69}{10}

The solutions to the quadratic equation are:

r = \frac{- 7 + 69}{10}, r = \frac{- 7 - 69}{10}

3 r^{2} - 4 r - 5 = 0

3 x^{2} + 7 x - 3 = 0

x^{2} + 40 = 4

100 x^{2} - 25 = 0

(x - 7)^{2} = 32

What are the solutions to the equation 2r^2+7r-1=-3r^2 ?
The solutions to the equation 2r^2+7r-1=-3r^2 are r=(-7+sqrt(69))/(10),r=(-7-sqrt(69))/(10)
Find the zeros of 2r^2+7r-1=-3r^2
The zeros of 2r^2+7r-1=-3r^2 are r=(-7+sqrt(69))/(10),r=(-7-sqrt(69))/(10)