What are the solutions to the equation v^2+11v=-28 ?

The solutions to the equation v^2+11v=-28 are v=-4,v=-7

Find the zeros of v^2+11v=-28

The zeros of v^2+11v=-28 are v=-4,v=-7

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v^{2} + 11 v = - 28

v = - 4, v = - 7

Solution steps

v^{2} + 11 v = - 28

Move

28

to the left side

v^{2} + 11 v + 28 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

1 1^{2} - 4 \cdot 1 \cdot 28 = 3

v_{1, 2} = \frac{- 11 \pm 3}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 11 + 3}{2 \cdot 1}, v_{2} = \frac{- 11 - 3}{2 \cdot 1}

v = \frac{- 11 + 3}{2 \cdot 1} : - 4

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

The solutions to the quadratic equation are:

v = - 4, v = - 7

0 = 32 + 16 t - 16 t^{2}

28 - 6 Q^{2} = 0

4 x^{2} = - 4 x - 16

3 g^{2} - 10 g - 13 = 0

x^{2} + 5^{2} + 2 = - 4

What are the solutions to the equation v^2+11v=-28 ?
The solutions to the equation v^2+11v=-28 are v=-4,v=-7
Find the zeros of v^2+11v=-28
The zeros of v^2+11v=-28 are v=-4,v=-7