What are the solutions to the equation (6x-7)^2-9(6x-7)-90=0 ?

The solutions to the equation (6x-7)^2-9(6x-7)-90=0 are x= 11/3 ,x= 1/6

Find the zeros of (6x-7)^2-9(6x-7)-90=0

The zeros of (6x-7)^2-9(6x-7)-90=0 are x= 11/3 ,x= 1/6

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(6 x - 7)^{2} - 9 (6 x - 7) - 90 = 0

x = \frac{11}{3}, x = \frac{1}{6}

Solution steps

(6 x - 7)^{2} - 9 (6 x - 7) - 90 = 0

Expand

(6 x - 7)^{2} - 9 (6 x - 7) - 90 : 36 x^{2} - 138 x + 22

36 x^{2} - 138 x + 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 138 ) \pm ( - 138 ) ^{2} - 4 \cdot 36 \cdot 22}{2 \cdot 36}

(- 138)^{2} - 4 \cdot 36 \cdot 22 = 126

x_{1, 2} = \frac{- ( - 138 ) \pm 126}{2 \cdot 36}

Separate the solutions

x_{1} = \frac{- ( - 138 ) + 126}{2 \cdot 36}, x_{2} = \frac{- ( - 138 ) - 126}{2 \cdot 36}

x = \frac{- ( - 138 ) + 126}{2 \cdot 36} : \frac{11}{3}

x = \frac{- ( - 138 ) - 126}{2 \cdot 36} : \frac{1}{6}

The solutions to the quadratic equation are:

x = \frac{11}{3}, x = \frac{1}{6}

- 8 x \geq 32

- 2 x^{2} + 8 x = 0

2 (x - 1)^{2} = 18

(\frac{3}{2})^{2} - 3 (\frac{3}{2}) + 4

y^{2} = y + 6

What are the solutions to the equation (6x-7)^2-9(6x-7)-90=0 ?
The solutions to the equation (6x-7)^2-9(6x-7)-90=0 are x= 11/3 ,x= 1/6
Find the zeros of (6x-7)^2-9(6x-7)-90=0
The zeros of (6x-7)^2-9(6x-7)-90=0 are x= 11/3 ,x= 1/6