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

The solutions to the equation 4x^2+2x-8=0 are x=(-1+sqrt(33))/4 ,x=-(1+sqrt(33))/4

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

The zeros of 4x^2+2x-8=0 are x=(-1+sqrt(33))/4 ,x=-(1+sqrt(33))/4

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

x = \frac{- 1 + 33}{4}, x = - \frac{1 + 33}{4}

Decimal Notation

x = 1.18614 \dots, x = - 1.68614 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 8) = 233

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- 1 + 33}{4}, x = - \frac{1 + 33}{4}

\frac{48}{- 16}

3 a^{2} b + 6 ab - 5 a^{3} b^{2} + 8 a^{2} b x + 4 a b^{2} m

\frac{6}{27}

(2 x + 3) (3 x - 1) \leq 0

x + 15 = 10

What are the solutions to the equation 4x^2+2x-8=0 ?
The solutions to the equation 4x^2+2x-8=0 are x=(-1+sqrt(33))/4 ,x=-(1+sqrt(33))/4
Find the zeros of 4x^2+2x-8=0
The zeros of 4x^2+2x-8=0 are x=(-1+sqrt(33))/4 ,x=-(1+sqrt(33))/4