What are the solutions to the equation v^2+3v-4=0 ?

The solutions to the equation v^2+3v-4=0 are v=1,v=-4

Find the zeros of v^2+3v-4=0

The zeros of v^2+3v-4=0 are v=1,v=-4

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v^{2} + 3 v - 4 = 0

v = 1, v = - 4

Solution steps

v^{2} + 3 v - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

v_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- 3 + 5}{2 \cdot 1}, v_{2} = \frac{- 3 - 5}{2 \cdot 1}

v = \frac{- 3 + 5}{2 \cdot 1} : 1

v = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

v = 1, v = - 4

6 x - 7 = 3 x + 2

∣ x + 4 ∣ \geq 7

\frac{( x - 3 )}{( x + 2 )} \leq 0

(x - 4)^{2} - 25 = 0

5 + c - 3 \leq 6

What are the solutions to the equation v^2+3v-4=0 ?
The solutions to the equation v^2+3v-4=0 are v=1,v=-4
Find the zeros of v^2+3v-4=0
The zeros of v^2+3v-4=0 are v=1,v=-4