What are the solutions to the equation 4x^2-6x-5=0 ?

The solutions to the equation 4x^2-6x-5=0 are x=(3+sqrt(29))/4 ,x=(3-sqrt(29))/4

Find the zeros of 4x^2-6x-5=0

The zeros of 4x^2-6x-5=0 are x=(3+sqrt(29))/4 ,x=(3-sqrt(29))/4

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4 x^{2} - 6 x - 5 = 0

x = \frac{3 + 29}{4}, x = \frac{3 - 29}{4}

Decimal Notation

x = 2.09629 \dots, x = - 0.59629 \dots

Solution steps

4 x^{2} - 6 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

(- 6)^{2} - 4 \cdot 4 (- 5) = 229

x_{1, 2} = \frac{- ( - 6 ) \pm 2 29}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2 29}{2 \cdot 4}, x_{2} = \frac{- ( - 6 ) - 2 29}{2 \cdot 4}

x = \frac{- ( - 6 ) + 2 29}{2 \cdot 4} : \frac{3 + 29}{4}

x = \frac{- ( - 6 ) - 2 29}{2 \cdot 4} : \frac{3 - 29}{4}

The solutions to the quadratic equation are:

x = \frac{3 + 29}{4}, x = \frac{3 - 29}{4}

\frac{2}{a ^{2} + 2 a + 1} + \frac{3}{a ^{2} - 1}

\frac{x}{2} + \frac{1}{3} = \frac{5}{6}

14 b^{2} + 39 b + 10

- 6 (1 + 2 x) \geq 6 (2 x - 1) + 2 x

x^{2} - 5 x = - 3 x + 24

What are the solutions to the equation 4x^2-6x-5=0 ?
The solutions to the equation 4x^2-6x-5=0 are x=(3+sqrt(29))/4 ,x=(3-sqrt(29))/4
Find the zeros of 4x^2-6x-5=0
The zeros of 4x^2-6x-5=0 are x=(3+sqrt(29))/4 ,x=(3-sqrt(29))/4