What are the solutions to the equation 19v^2+5v-18=11v+10v^2 ?

The solutions to the equation 19v^2+5v-18=11v+10v^2 are v=(1+sqrt(19))/3 ,v=(1-sqrt(19))/3

Find the zeros of 19v^2+5v-18=11v+10v^2

The zeros of 19v^2+5v-18=11v+10v^2 are v=(1+sqrt(19))/3 ,v=(1-sqrt(19))/3

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19 v^{2} + 5 v - 18 = 11 v + 10 v^{2}

v = \frac{1 + 19}{3}, v = \frac{1 - 19}{3}

Decimal

v = 1.78629 \dots, v = - 1.11963 \dots

Solution steps

19 v^{2} + 5 v - 18 = 11 v + 10 v^{2}

Move

10 v^{2}

to the left side

9 v^{2} + 5 v - 18 = 11 v

Move

11 v

to the left side

9 v^{2} - 6 v - 18 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 9 ( - 18 )}{2 \cdot 9}

(- 6)^{2} - 4 \cdot 9 (- 18) = 619

v_{1, 2} = \frac{- ( - 6 ) \pm 6 19}{2 \cdot 9}

Separate the solutions

v_{1} = \frac{- ( - 6 ) + 6 19}{2 \cdot 9}, v_{2} = \frac{- ( - 6 ) - 6 19}{2 \cdot 9}

v = \frac{- ( - 6 ) + 6 19}{2 \cdot 9} : \frac{1 + 19}{3}

v = \frac{- ( - 6 ) - 6 19}{2 \cdot 9} : \frac{1 - 19}{3}

The solutions to the quadratic equation are:

v = \frac{1 + 19}{3}, v = \frac{1 - 19}{3}

4 (1 - x) \geq 3 (x - 1)

8 (x - 4) = 32

2 (- \frac{1}{2}) + 1

\frac{720}{18088} \cdot 100

338

What are the solutions to the equation 19v^2+5v-18=11v+10v^2 ?
The solutions to the equation 19v^2+5v-18=11v+10v^2 are v=(1+sqrt(19))/3 ,v=(1-sqrt(19))/3
Find the zeros of 19v^2+5v-18=11v+10v^2
The zeros of 19v^2+5v-18=11v+10v^2 are v=(1+sqrt(19))/3 ,v=(1-sqrt(19))/3