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

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

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

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

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

v = 1, v = - 3

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} : 1

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

The solutions to the quadratic equation are:

v = 1, v = - 3

6 b p^{3} - 27 p^{2} - 10 b pq + 45 q

y^{2} + 3 y - 4

x^{2} = 4 x + 32

x^{2} + 6 x + 9 = x + 45

\frac{3 x ^{- 2}}{y ^{5} w ^{- 2}}

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