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

The solutions to the equation 4x^2+7x-4=0 are x=(-7+sqrt(113))/8 ,x=(-7-sqrt(113))/8

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

The zeros of 4x^2+7x-4=0 are x=(-7+sqrt(113))/8 ,x=(-7-sqrt(113))/8

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

x = \frac{- 7 + 113}{8}, x = \frac{- 7 - 113}{8}

Decimal Notation

x = 0.45376 \dots, x = - 2.20376 \dots

Solution steps

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

Solve with the quadratic formula

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

7^{2} - 4 \cdot 4 (- 4) = 113

x_{1, 2} = \frac{- 7 \pm 113}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 7 + 113}{2 \cdot 4}, x_{2} = \frac{- 7 - 113}{2 \cdot 4}

x = \frac{- 7 + 113}{2 \cdot 4} : \frac{- 7 + 113}{8}

x = \frac{- 7 - 113}{2 \cdot 4} : \frac{- 7 - 113}{8}

The solutions to the quadratic equation are:

x = \frac{- 7 + 113}{8}, x = \frac{- 7 - 113}{8}

n^{2} = 18 n + 40

x, x^{2} + 3 x + 4 = m x

8 x^{2} - 50 = 0

x, z = x^{2} e^{- y}

3 r^{2} - 12 r - 77 = 5

What are the solutions to the equation 4x^2+7x-4=0 ?
The solutions to the equation 4x^2+7x-4=0 are x=(-7+sqrt(113))/8 ,x=(-7-sqrt(113))/8
Find the zeros of 4x^2+7x-4=0
The zeros of 4x^2+7x-4=0 are x=(-7+sqrt(113))/8 ,x=(-7-sqrt(113))/8