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

The solutions to the equation (x-7)^2+(x-7)=6 are x=9,x=4

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

The zeros of (x-7)^2+(x-7)=6 are x=9,x=4

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

x = 9, x = 4

Solution steps

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

Expand

(x - 7)^{2} + (x - 7) : x^{2} - 13 x + 42

x^{2} - 13 x + 42 = 6

Move

6

to the left side

x^{2} - 13 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 13)^{2} - 4 \cdot 1 \cdot 36 = 5

x_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 1}

x = \frac{- ( - 13 ) + 5}{2 \cdot 1} : 9

x = \frac{- ( - 13 ) - 5}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 9, x = 4

a x^{2} - b x - c = 0

- 3 x^{2} - 12 x = 4

2 \cdot 3 + 12 = x^{2} - 13

(2 x - 7) (x + 9) = 0

x, 3 y - 6 = 12 x^{2} - 6 y^{3}

What are the solutions to the equation (x-7)^2+(x-7)=6 ?
The solutions to the equation (x-7)^2+(x-7)=6 are x=9,x=4
Find the zeros of (x-7)^2+(x-7)=6
The zeros of (x-7)^2+(x-7)=6 are x=9,x=4