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 = 1.39564 \dots, x = - 0.89564 \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}

\frac{x ^{- 9} y ^{6}}{x ^{4} y ^{10}}

\frac{5}{6} - \frac{2}{3}

lo g_{4} (0.125)

0 = 88 t - 16 t^{2}

(\frac{e}{s - 2})^{- 2 s}

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