What are the solutions to the equation 4v^2+7v-7=0 ?

The solutions to the equation 4v^2+7v-7=0 are v=(-7+sqrt(161))/8 ,v=(-7-sqrt(161))/8

Find the zeros of 4v^2+7v-7=0

The zeros of 4v^2+7v-7=0 are v=(-7+sqrt(161))/8 ,v=(-7-sqrt(161))/8

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4 v^{2} + 7 v - 7 = 0

v = \frac{- 7 + 161}{8}, v = \frac{- 7 - 161}{8}

Decimal Notation

v = 0.71107 \dots, v = - 2.46107 \dots

Solution steps

4 v^{2} + 7 v - 7 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 7 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 7) = 161

v_{1, 2} = \frac{- 7 \pm 161}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- 7 + 161}{2 \cdot 4}, v_{2} = \frac{- 7 - 161}{2 \cdot 4}

v = \frac{- 7 + 161}{2 \cdot 4} : \frac{- 7 + 161}{8}

v = \frac{- 7 - 161}{2 \cdot 4} : \frac{- 7 - 161}{8}

The solutions to the quadratic equation are:

v = \frac{- 7 + 161}{8}, v = \frac{- 7 - 161}{8}

12 x - 3 x^{2} = 0

3 x^{2} - 12 x + 15 = 0

x^{2} + 18 x + 57 = 12

5 x^{2} - 57 x = 3 x - 120

(2 - x)^{2} = 0

What are the solutions to the equation 4v^2+7v-7=0 ?
The solutions to the equation 4v^2+7v-7=0 are v=(-7+sqrt(161))/8 ,v=(-7-sqrt(161))/8
Find the zeros of 4v^2+7v-7=0
The zeros of 4v^2+7v-7=0 are v=(-7+sqrt(161))/8 ,v=(-7-sqrt(161))/8