What are the solutions to the equation 4x^2-x-1=5 ?

The solutions to the equation 4x^2-x-1=5 are x=(1+sqrt(97))/8 ,x=(1-sqrt(97))/8

Find the zeros of 4x^2-x-1=5

The zeros of 4x^2-x-1=5 are x=(1+sqrt(97))/8 ,x=(1-sqrt(97))/8

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4 x^{2} - x - 1 = 5

x = \frac{1 + 97}{8}, x = \frac{1 - 97}{8}

Decimal Notation

x = 1.35610 \dots, x = - 1.10610 \dots

Solution steps

4 x^{2} - x - 1 = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 4 (- 6) = 97

x_{1, 2} = \frac{- ( - 1 ) \pm 97}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 97}{2 \cdot 4}, x_{2} = \frac{- ( - 1 ) - 97}{2 \cdot 4}

x = \frac{- ( - 1 ) + 97}{2 \cdot 4} : \frac{1 + 97}{8}

x = \frac{- ( - 1 ) - 97}{2 \cdot 4} : \frac{1 - 97}{8}

The solutions to the quadratic equation are:

x = \frac{1 + 97}{8}, x = \frac{1 - 97}{8}

36 x^{2} - 49 = 0

2 x^{2} - 5 x = 6

40 = 30 t - 5 t^{2}

f, f (1 - x) = x^{2} - 2 x

4 x^{2} + 8 x + 3 = a (x + b)^{2} + e

What are the solutions to the equation 4x^2-x-1=5 ?
The solutions to the equation 4x^2-x-1=5 are x=(1+sqrt(97))/8 ,x=(1-sqrt(97))/8
Find the zeros of 4x^2-x-1=5
The zeros of 4x^2-x-1=5 are x=(1+sqrt(97))/8 ,x=(1-sqrt(97))/8