What are the solutions to the equation 4x^2+2x-5=0 ?

The solutions to the equation 4x^2+2x-5=0 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4

Find the zeros of 4x^2+2x-5=0

The zeros of 4x^2+2x-5=0 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4

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4 x^{2} + 2 x - 5 = 0

x = \frac{- 1 + 21}{4}, x = - \frac{1 + 21}{4}

Decimal Notation

x = 0.89564 \dots, x = - 1.39564 \dots

Solution steps

4 x^{2} + 2 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

2^{2} - 4 \cdot 4 (- 5) = 221

x_{1, 2} = \frac{- 2 \pm 2 21}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 2 + 2 21}{2 \cdot 4}, x_{2} = \frac{- 2 - 2 21}{2 \cdot 4}

x = \frac{- 2 + 2 21}{2 \cdot 4} : \frac{- 1 + 21}{4}

x = \frac{- 2 - 2 21}{2 \cdot 4} : - \frac{1 + 21}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{4}, x = - \frac{1 + 21}{4}

x^{2} + 40 x + 2000 = 0

176 x = 121 + 64 x^{2}

3 x^{2} - 300 = 0

(x + 8)^{2} - 43 = - 38

z^{2} - 17 z + 12 = - 5 z

What are the solutions to the equation 4x^2+2x-5=0 ?
The solutions to the equation 4x^2+2x-5=0 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4
Find the zeros of 4x^2+2x-5=0
The zeros of 4x^2+2x-5=0 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4