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

The solutions to the equation (x^2-4x)+x-4=0 are x=4,x=-1

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

The zeros of (x^2-4x)+x-4=0 are x=4,x=-1

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

x = 4, x = - 1

Solution steps

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

Refine

x^{2} - 3 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

x = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

x = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 4, x = - 1

- 2 x - 3 \leq - 5

3 x^{2} + 6 x + 15 = 10

in (x^{2})

6 (2 x - 3) + 4 = 16 - (- 18)

r^{2} - 16 r + 48

What are the solutions to the equation (x^2-4x)+x-4=0 ?
The solutions to the equation (x^2-4x)+x-4=0 are x=4,x=-1
Find the zeros of (x^2-4x)+x-4=0
The zeros of (x^2-4x)+x-4=0 are x=4,x=-1