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}

Decimal Notation

x = 2.85078 \dots, x = - 0.35078 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 10)^{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}

n^{2} - 10 = 26

y^{2} - 2 y - 48 = 0

(4 x - 1)^{2} + 8 x = 4 x + 1

- 20 - 3 x + 3 x^{2} = 0

(2 x - 5)^{2} - 81 = 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