What are the solutions to the equation 3v^2+17v+24=5 ?

The solutions to the equation 3v^2+17v+24=5 are v=(-17+sqrt(61))/6 ,v=(-17-sqrt(61))/6

Find the zeros of 3v^2+17v+24=5

The zeros of 3v^2+17v+24=5 are v=(-17+sqrt(61))/6 ,v=(-17-sqrt(61))/6

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3 v^{2} + 17 v + 24 = 5

v = \frac{- 17 + 61}{6}, v = \frac{- 17 - 61}{6}

Solution steps

3 v^{2} + 17 v + 24 = 5

Move

5

to the left side

3 v^{2} + 17 v + 19 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 3 \cdot 19}{2 \cdot 3}

1 7^{2} - 4 \cdot 3 \cdot 19 = 61

v_{1, 2} = \frac{- 17 \pm 61}{2 \cdot 3}

Separate the solutions

v_{1} = \frac{- 17 + 61}{2 \cdot 3}, v_{2} = \frac{- 17 - 61}{2 \cdot 3}

v = \frac{- 17 + 61}{2 \cdot 3} : \frac{- 17 + 61}{6}

v = \frac{- 17 - 61}{2 \cdot 3} : \frac{- 17 - 61}{6}

The solutions to the quadratic equation are:

v = \frac{- 17 + 61}{6}, v = \frac{- 17 - 61}{6}

\frac{26 !}{( 26 - 4 ) !}

\frac{z}{7} - 6 \geq - 5

5 ∣ 2 x - 3 ∣ = 15

2 x (x - 3) + 9 (x - 3)

\frac{- x}{3} + \frac{x}{2} = - 1

What are the solutions to the equation 3v^2+17v+24=5 ?
The solutions to the equation 3v^2+17v+24=5 are v=(-17+sqrt(61))/6 ,v=(-17-sqrt(61))/6
Find the zeros of 3v^2+17v+24=5
The zeros of 3v^2+17v+24=5 are v=(-17+sqrt(61))/6 ,v=(-17-sqrt(61))/6