What are the solutions to the equation v^2-3v-28=0 ?

The solutions to the equation v^2-3v-28=0 are v=7,v=-4

Find the zeros of v^2-3v-28=0

The zeros of v^2-3v-28=0 are v=7,v=-4

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v^{2} - 3 v - 28 = 0

v = 7, v = - 4

Solution steps

v^{2} - 3 v - 28 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 28) = 11

v_{1, 2} = \frac{- ( - 3 ) \pm 11}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 3 ) + 11}{2 \cdot 1}, v_{2} = \frac{- ( - 3 ) - 11}{2 \cdot 1}

v = \frac{- ( - 3 ) + 11}{2 \cdot 1} : 7

v = \frac{- ( - 3 ) - 11}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

v = 7, v = - 4

4 r - 5 = - \frac{1}{2} - 4 (r + 1)

(3 y + 2)^{2} + 10 = - 18

1 = 2

\frac{k ^{2} + 9 k - 10}{34} \cdot \frac{2}{k ^{2} - 1}

3 x + 1 = x + 8

What are the solutions to the equation v^2-3v-28=0 ?
The solutions to the equation v^2-3v-28=0 are v=7,v=-4
Find the zeros of v^2-3v-28=0
The zeros of v^2-3v-28=0 are v=7,v=-4