What are the solutions to the equation v^2-2v-35=0 ?

The solutions to the equation v^2-2v-35=0 are v=7,v=-5

Find the zeros of v^2-2v-35=0

The zeros of v^2-2v-35=0 are v=7,v=-5

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v^{2} - 2 v - 35 = 0

v = 7, v = - 5

Solution steps

v^{2} - 2 v - 35 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 35 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 35) = 12

v_{1, 2} = \frac{- ( - 2 ) \pm 12}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 2 ) + 12}{2 \cdot 1}, v_{2} = \frac{- ( - 2 ) - 12}{2 \cdot 1}

v = \frac{- ( - 2 ) + 12}{2 \cdot 1} : 7

v = \frac{- ( - 2 ) - 12}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

v = 7, v = - 5

- 108 = - 16 t^{2} - 22 t

2 x^{2} - 10 = x^{2} + 15

w^{2} = 5 (w - 4)

6 x^{2} - 7 x + 6 = 0

- 5 (3 x + 15) (x - 6) = 0

What are the solutions to the equation v^2-2v-35=0 ?
The solutions to the equation v^2-2v-35=0 are v=7,v=-5
Find the zeros of v^2-2v-35=0
The zeros of v^2-2v-35=0 are v=7,v=-5