What are the solutions to the equation (2-x)^2+(2-x)-20=0 ?

The solutions to the equation (2-x)^2+(2-x)-20=0 are x=7,x=-2

Find the zeros of (2-x)^2+(2-x)-20=0

The zeros of (2-x)^2+(2-x)-20=0 are x=7,x=-2

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(2 - x)^{2} + (2 - x) - 20 = 0

x = 7, x = - 2

Solution steps

(2 - x)^{2} + (2 - x) - 20 = 0

Expand

(2 - x)^{2} + (2 - x) - 20 : x^{2} - 5 x - 14

x^{2} - 5 x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 14 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 14) = 9

x_{1, 2} = \frac{- ( - 5 ) \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 9}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 9}{2 \cdot 1}

x = \frac{- ( - 5 ) + 9}{2 \cdot 1} : 7

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

The solutions to the quadratic equation are:

x = 7, x = - 2

2 x (x + 4) = (x + 3) (x - 3)

z^{2} - 5 \cdot 25 = 1 0^{2}

(2 x + 3) = x (1 - 3 x)

(t + 4) (t + 7) = 0

6 x^{2} - 31 x + 5 = 0

What are the solutions to the equation (2-x)^2+(2-x)-20=0 ?
The solutions to the equation (2-x)^2+(2-x)-20=0 are x=7,x=-2
Find the zeros of (2-x)^2+(2-x)-20=0
The zeros of (2-x)^2+(2-x)-20=0 are x=7,x=-2