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

The solutions to the equation (x-5)^2+4(x-5)-5=0 are x=6,x=0

Find the zeros of (x-5)^2+4(x-5)-5=0

The zeros of (x-5)^2+4(x-5)-5=0 are x=6,x=0

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(x - 5)^{2} + 4 (x - 5) - 5 = 0

x = 6, x = 0

Solution steps

(x - 5)^{2} + 4 (x - 5) - 5 = 0

Expand

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

x^{2} - 6 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 0 = 6

x_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 6}{2 \cdot 1}

x = \frac{- ( - 6 ) + 6}{2 \cdot 1} : 6

x = \frac{- ( - 6 ) - 6}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 6, x = 0

\frac{5 π}{6} + \frac{2 π}{3}

5 x - 6 = 8

9 x^{2} - 7 = 18

\frac{- 9 i}{8 i}

80 y^{2} - 5 z^{2}

What are the solutions to the equation (x-5)^2+4(x-5)-5=0 ?
The solutions to the equation (x-5)^2+4(x-5)-5=0 are x=6,x=0
Find the zeros of (x-5)^2+4(x-5)-5=0
The zeros of (x-5)^2+4(x-5)-5=0 are x=6,x=0