What are the solutions to the equation (x-5)^2+(x-5)=0 ?

The solutions to the equation (x-5)^2+(x-5)=0 are x=5,x=4

Find the zeros of (x-5)^2+(x-5)=0

The zeros of (x-5)^2+(x-5)=0 are x=5,x=4

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

x = 5, x = 4

Solution steps

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

Expand

(x - 5)^{2} + (x - 5) : x^{2} - 9 x + 20

x^{2} - 9 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 20 = 1

x_{1, 2} = \frac{- ( - 9 ) \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 1}{2 \cdot 1}, x_{2} = \frac{- ( - 9 ) - 1}{2 \cdot 1}

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

x = \frac{- ( - 9 ) - 1}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

x = 5, x = 4

x, - x^{2} - 3 y^{3} = 5

2 x^{2} = 20 - 3 x^{2}

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

x^{2} + 18 x + 63 = 0

3 (3 x + 2) - 6 x - 5 = 4 x^{2}

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