What are the solutions to the equation v^2+4v-21=0 ?

The solutions to the equation v^2+4v-21=0 are v=3,v=-7

Find the zeros of v^2+4v-21=0

The zeros of v^2+4v-21=0 are v=3,v=-7

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v^{2} + 4 v - 21 = 0

v = 3, v = - 7

Solution steps

v^{2} + 4 v - 21 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 21 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 21) = 10

v_{1, 2} = \frac{- 4 \pm 10}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 4 + 10}{2 \cdot 1}, v_{2} = \frac{- 4 - 10}{2 \cdot 1}

v = \frac{- 4 + 10}{2 \cdot 1} : 3

v = \frac{- 4 - 10}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

v = 3, v = - 7

3 x + 2 (4 x - 4) = 3

8 x^{2} + 14 x = - 3

2 x = 10

- \frac{1}{27} (- 3)

∣ x + 8 ∣ = 2

What are the solutions to the equation v^2+4v-21=0 ?
The solutions to the equation v^2+4v-21=0 are v=3,v=-7
Find the zeros of v^2+4v-21=0
The zeros of v^2+4v-21=0 are v=3,v=-7