What are the solutions to the equation 4x^2+10x-4=0 ?

The solutions to the equation 4x^2+10x-4=0 are x=(-5+sqrt(41))/4 ,x=-(5+sqrt(41))/4

Find the zeros of 4x^2+10x-4=0

The zeros of 4x^2+10x-4=0 are x=(-5+sqrt(41))/4 ,x=-(5+sqrt(41))/4

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

x = \frac{- 5 + 41}{4}, x = - \frac{5 + 41}{4}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

1 0^{2} - 4 \cdot 4 (- 4) = 241

x_{1, 2} = \frac{- 10 \pm 2 41}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 10 + 2 41}{2 \cdot 4}, x_{2} = \frac{- 10 - 2 41}{2 \cdot 4}

x = \frac{- 10 + 2 41}{2 \cdot 4} : \frac{- 5 + 41}{4}

x = \frac{- 10 - 2 41}{2 \cdot 4} : - \frac{5 + 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 41}{4}, x = - \frac{5 + 41}{4}

(x + 5) 2 + 6 = 2 (x + 4) (x + 2)

16 x^{2} = 11

4 x^{2} - 8 x + 7

4 x^{2} - 8 x + 2

0.01 y^{2} - 0.02 y - 0.01 = 0

What are the solutions to the equation 4x^2+10x-4=0 ?
The solutions to the equation 4x^2+10x-4=0 are x=(-5+sqrt(41))/4 ,x=-(5+sqrt(41))/4
Find the zeros of 4x^2+10x-4=0
The zeros of 4x^2+10x-4=0 are x=(-5+sqrt(41))/4 ,x=-(5+sqrt(41))/4