What are the solutions to the equation 4x^2+31x-8=0 ?

The solutions to the equation 4x^2+31x-8=0 are x= 1/4 ,x=-8

Find the zeros of 4x^2+31x-8=0

The zeros of 4x^2+31x-8=0 are x= 1/4 ,x=-8

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4 x^{2} + 31 x - 8 = 0

x = \frac{1}{4}, x = - 8

Decimal Notation

x = 0.25, x = - 8

Solution steps

4 x^{2} + 31 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 31 \pm 3 1 ^{2} - 4 \cdot 4 ( - 8 )}{2 \cdot 4}

3 1^{2} - 4 \cdot 4 (- 8) = 33

x_{1, 2} = \frac{- 31 \pm 33}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 31 + 33}{2 \cdot 4}, x_{2} = \frac{- 31 - 33}{2 \cdot 4}

x = \frac{- 31 + 33}{2 \cdot 4} : \frac{1}{4}

x = \frac{- 31 - 33}{2 \cdot 4} : - 8

The solutions to the quadratic equation are:

x = \frac{1}{4}, x = - 8

20 x^{2} + 10 x + 5 = 11 x + 6

(3 + x) \frac{2}{5} + 6 (3 - x) \frac{2}{5} = 5 (9 - x^{2}) \frac{1}{5}

x^{2} - 10 = 71

x^{2} + 5 x - 22 = 6 x + 2

x, x^{2} + y^{2} = x^{2} y^{2}

What are the solutions to the equation 4x^2+31x-8=0 ?
The solutions to the equation 4x^2+31x-8=0 are x= 1/4 ,x=-8
Find the zeros of 4x^2+31x-8=0
The zeros of 4x^2+31x-8=0 are x= 1/4 ,x=-8