What are the solutions to the equation v^2+2v+4=3 ?

The solutions to the equation v^2+2v+4=3 are v=-1

Find the zeros of v^2+2v+4=3

The zeros of v^2+2v+4=3 are v=-1

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v^{2} + 2 v + 4 = 3

v = - 1

Solution steps

v^{2} + 2 v + 4 = 3

Move

3

to the left side

v^{2} + 2 v + 1 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot 1 = 0

v_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

v = \frac{- 2}{2 \cdot 1}

\frac{- 2}{2 \cdot 1} = - 1

v = - 1

The solution to the quadratic equation is:

v = - 1

2 x^{2} + 24 = 16 x

y, x^{2} - 8 x y + y^{2} = 8

a x^{2} + b x - 1

x^{2} = 2 x - 2

0 = 3 (x - 4)^{2}

What are the solutions to the equation v^2+2v+4=3 ?
The solutions to the equation v^2+2v+4=3 are v=-1
Find the zeros of v^2+2v+4=3
The zeros of v^2+2v+4=3 are v=-1