What are the solutions to the equation x=1(5-6x-1)^2 ?

The solutions to the equation x=1(5-6x-1)^2 are x=(49+sqrt(97))/(72),x=(49-sqrt(97))/(72)

Find the zeros of x=1(5-6x-1)^2

The zeros of x=1(5-6x-1)^2 are x=(49+sqrt(97))/(72),x=(49-sqrt(97))/(72)

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x = 1 (5 - 6 x - 1)^{2}

x = \frac{49 + 97}{72}, x = \frac{49 - 97}{72}

Decimal Notation

x = 0.81734 \dots, x = 0.54376 \dots

Solution steps

x = 1 \cdot (5 - 6 x - 1)^{2}

Expand

1 \cdot (5 - 6 x - 1)^{2} : 36 x^{2} - 48 x + 16

x = 36 x^{2} - 48 x + 16

Switch sides

36 x^{2} - 48 x + 16 = x

Move

x

to the left side

36 x^{2} - 49 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 49 ) \pm ( - 49 ) ^{2} - 4 \cdot 36 \cdot 16}{2 \cdot 36}

(- 49)^{2} - 4 \cdot 36 \cdot 16 = 97

x_{1, 2} = \frac{- ( - 49 ) \pm 97}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- ( - 49 ) + 97}{2 \cdot 36}, x_{2} = \frac{- ( - 49 ) - 97}{2 \cdot 36}

x = \frac{- ( - 49 ) + 97}{2 \cdot 36} : \frac{49 + 97}{72}

x = \frac{- ( - 49 ) - 97}{2 \cdot 36} : \frac{49 - 97}{72}

The solutions to the quadratic equation are:

x = \frac{49 + 97}{72}, x = \frac{49 - 97}{72}

(k + 2) x^{2} - 5 x + \frac{5}{2} = 0

x, x^{2} + y^{2} - 6 x - 10 y + 34 = 0

3.57 x^{2} - 160 x + 500 = 0

(x + 6)^{2} = 50

u^{2} + 6 u - 7 = 0

What are the solutions to the equation x=1(5-6x-1)^2 ?
The solutions to the equation x=1(5-6x-1)^2 are x=(49+sqrt(97))/(72),x=(49-sqrt(97))/(72)
Find the zeros of x=1(5-6x-1)^2
The zeros of x=1(5-6x-1)^2 are x=(49+sqrt(97))/(72),x=(49-sqrt(97))/(72)