What are the solutions to the equation (6x-7)^2+12(6x-7)+36=0 ?

The solutions to the equation (6x-7)^2+12(6x-7)+36=0 are x= 1/6

Find the zeros of (6x-7)^2+12(6x-7)+36=0

The zeros of (6x-7)^2+12(6x-7)+36=0 are x= 1/6

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(6 x - 7)^{2} + 12 (6 x - 7) + 36 = 0

x = \frac{1}{6}

Decimal Notation

x = 0.16666 \dots

Solution steps

(6 x - 7)^{2} + 12 (6 x - 7) + 36 = 0

Expand

(6 x - 7)^{2} + 12 (6 x - 7) + 36 : 36 x^{2} - 12 x + 1

36 x^{2} - 12 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 36 \cdot 1}{2 \cdot 36}

(- 12)^{2} - 4 \cdot 36 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 36}

x = \frac{- ( - 12 )}{2 \cdot 36}

\frac{- ( - 12 )}{2 \cdot 36} = \frac{1}{6}

x = \frac{1}{6}

The solution to the quadratic equation is:

x = \frac{1}{6}

x^{2} + 5 x - 14 = - 8

x^{2} + 6 x + 7 = - 2 x

4 x^{2} + 16 x + 19

x^{2} + 9 = 125

x^{2} + 5 x + 30 = 0

What are the solutions to the equation (6x-7)^2+12(6x-7)+36=0 ?
The solutions to the equation (6x-7)^2+12(6x-7)+36=0 are x= 1/6
Find the zeros of (6x-7)^2+12(6x-7)+36=0
The zeros of (6x-7)^2+12(6x-7)+36=0 are x= 1/6