What are the solutions to the equation v^2-2v-3=0 ?

The solutions to the equation v^2-2v-3=0 are v=3,v=-1

Find the zeros of v^2-2v-3=0

The zeros of v^2-2v-3=0 are v=3,v=-1

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

v = 3, v = - 1

Solution steps

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

0.42 (0.6 - x) = x^{2}

∣ 7 x + 9 ∣ < - 4

(\frac{1}{3})^{4 x - 5} > \frac{1}{27}

∣ 1 - x ∣ < 2

6 x^{2} + 10 x = - 1

What are the solutions to the equation v^2-2v-3=0 ?
The solutions to the equation v^2-2v-3=0 are v=3,v=-1
Find the zeros of v^2-2v-3=0
The zeros of v^2-2v-3=0 are v=3,v=-1