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

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

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

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

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

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

Decimal Notation

x = 0.71107 \dots, x = - 2.46107 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x^{2} - 5 x + 1

(6 x - 3) (2 x + 4) = 3

4 x^{2} - 2 x + 10 = 0

x^{2} + 9 x = 27

2 x^{2} - 10 x + 9 = 0

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