What are the solutions to the equation v^2+11v+24=0 ?

The solutions to the equation v^2+11v+24=0 are v=-3,v=-8

Find the zeros of v^2+11v+24=0

The zeros of v^2+11v+24=0 are v=-3,v=-8

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v^{2} + 11 v + 24 = 0

v = - 3, v = - 8

Solution steps

v^{2} + 11 v + 24 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 24 = 5

v_{1, 2} = \frac{- 11 \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 11 + 5}{2 \cdot 1}, v_{2} = \frac{- 11 - 5}{2 \cdot 1}

v = \frac{- 11 + 5}{2 \cdot 1} : - 3

v = \frac{- 11 - 5}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

v = - 3, v = - 8

y, x^{2} y^{2} + 3 y = 4 x

2 x^{2} + \frac{3}{7} x - 8 = 0

(x - 4)^{2} - 121 = 0

x^{2} = 12 x

x, z = 4 - x^{2} - y^{2}

What are the solutions to the equation v^2+11v+24=0 ?
The solutions to the equation v^2+11v+24=0 are v=-3,v=-8
Find the zeros of v^2+11v+24=0
The zeros of v^2+11v+24=0 are v=-3,v=-8