What are the solutions to the equation 19v^2-v-2=4v^2 ?

The solutions to the equation 19v^2-v-2=4v^2 are v= 2/5 ,v=-1/3

Find the zeros of 19v^2-v-2=4v^2

The zeros of 19v^2-v-2=4v^2 are v= 2/5 ,v=-1/3

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19 v^{2} - v - 2 = 4 v^{2}

v = \frac{2}{5}, v = - \frac{1}{3}

Decimal Notation

v = 0.4, v = - 0.33333 \dots

Solution steps

19 v^{2} - v - 2 = 4 v^{2}

Move

4 v^{2}

to the left side

15 v^{2} - v - 2 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 15 ( - 2 )}{2 \cdot 15}

(- 1)^{2} - 4 \cdot 15 (- 2) = 11

v_{1, 2} = \frac{- ( - 1 ) \pm 11}{2 \cdot 15}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 11}{2 \cdot 15}, v_{2} = \frac{- ( - 1 ) - 11}{2 \cdot 15}

v = \frac{- ( - 1 ) + 11}{2 \cdot 15} : \frac{2}{5}

v = \frac{- ( - 1 ) - 11}{2 \cdot 15} : - \frac{1}{3}

The solutions to the quadratic equation are:

v = \frac{2}{5}, v = - \frac{1}{3}

2 x (x - 2) = 200

(1 - x)^{2} = e^{3}

- k^{2} - 3 k - 2 = - 5 k^{2}

x^{2} = 10 - 3 x

9 x^{2} - 12 x + 13 = 0

What are the solutions to the equation 19v^2-v-2=4v^2 ?
The solutions to the equation 19v^2-v-2=4v^2 are v= 2/5 ,v=-1/3
Find the zeros of 19v^2-v-2=4v^2
The zeros of 19v^2-v-2=4v^2 are v= 2/5 ,v=-1/3