What are the solutions to the equation (13x-1)^2-3(13x-1)-4=0 ?

The solutions to the equation (13x-1)^2-3(13x-1)-4=0 are x= 5/13 ,x=0

Find the zeros of (13x-1)^2-3(13x-1)-4=0

The zeros of (13x-1)^2-3(13x-1)-4=0 are x= 5/13 ,x=0

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(13 x - 1)^{2} - 3 (13 x - 1) - 4 = 0

x = \frac{5}{13}, x = 0

Decimal Notation

x = 0.38461 \dots, x = 0

Solution steps

(13 x - 1)^{2} - 3 (13 x - 1) - 4 = 0

Expand

(13 x - 1)^{2} - 3 (13 x - 1) - 4 : 169 x^{2} - 65 x

169 x^{2} - 65 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 65 ) \pm ( - 65 ) ^{2} - 4 \cdot 169 \cdot 0}{2 \cdot 169}

(- 65)^{2} - 4 \cdot 169 \cdot 0 = 65

x_{1, 2} = \frac{- ( - 65 ) \pm 65}{2 \cdot 169}

Separate the solutions

x_{1} = \frac{- ( - 65 ) + 65}{2 \cdot 169}, x_{2} = \frac{- ( - 65 ) - 65}{2 \cdot 169}

x = \frac{- ( - 65 ) + 65}{2 \cdot 169} : \frac{5}{13}

x = \frac{- ( - 65 ) - 65}{2 \cdot 169} : 0

The solutions to the quadratic equation are:

x = \frac{5}{13}, x = 0

\frac{1}{( 1 ) ^{2} + 1}

2 x^{2} - 4 x = 30

\frac{x ^{3} + y ^{3}}{x ^{2} - y ^{2}}

\frac{3}{2} x + 2 = \frac{1}{2} - \frac{1}{2} x

\frac{x ^{2} y ^{3}}{x y ^{4}}

What are the solutions to the equation (13x-1)^2-3(13x-1)-4=0 ?
The solutions to the equation (13x-1)^2-3(13x-1)-4=0 are x= 5/13 ,x=0
Find the zeros of (13x-1)^2-3(13x-1)-4=0
The zeros of (13x-1)^2-3(13x-1)-4=0 are x= 5/13 ,x=0