What are the solutions to the equation 5v^2+7v-3=-v^2 ?

The solutions to the equation 5v^2+7v-3=-v^2 are v= 1/3 ,v=-3/2

Find the zeros of 5v^2+7v-3=-v^2

The zeros of 5v^2+7v-3=-v^2 are v= 1/3 ,v=-3/2

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5 v^{2} + 7 v - 3 = - v^{2}

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

Decimal Notation

v = 0.33333 \dots, v = - 1.5

Solution steps

5 v^{2} + 7 v - 3 = - v^{2}

Move

v^{2}

to the left side

6 v^{2} + 7 v - 3 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 6 ( - 3 )}{2 \cdot 6}

7^{2} - 4 \cdot 6 (- 3) = 11

v_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 6}

Separate the solutions

v_{1} = \frac{- 7 + 11}{2 \cdot 6}, v_{2} = \frac{- 7 - 11}{2 \cdot 6}

v = \frac{- 7 + 11}{2 \cdot 6} : \frac{1}{3}

v = \frac{- 7 - 11}{2 \cdot 6} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

lo g_{5} (125)

6 y^{2} + 6 y - 12 = 0

6 m^{4} + 27 m^{3} + 6 m^{2} + 27 m

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} - \frac{n ( n + 1 )}{2}

- \frac{1}{5} \cdot 10

What are the solutions to the equation 5v^2+7v-3=-v^2 ?
The solutions to the equation 5v^2+7v-3=-v^2 are v= 1/3 ,v=-3/2
Find the zeros of 5v^2+7v-3=-v^2
The zeros of 5v^2+7v-3=-v^2 are v= 1/3 ,v=-3/2