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

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

Find the zeros of v^2-2v=3

The zeros of v^2-2v=3 are v=3,v=-1

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v^{2} - 2 v = 3

v = 3, v = - 1

Solution steps

v^{2} - 2 v = 3

Move

3

to the left side

v^{2} - 2 v - 3 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

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

Separate the solutions

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

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

v = \frac{- ( - 2 ) - 4}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

v = 3, v = - 1

2 (x - 3)^{2} + (7 x - 21) - 15 = 0

a^{2} - 9 a + 2 = 0

x^{2} + (1 - 2 k) x + (k^{2} - k - 2) = 0

x^{2} - 13 x - 420 = 0

x, p (x) = y (x^{2} + 2) - x (y^{2} + 2)

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