What are the solutions to the equation x^2+(-x+3)^2-4x-21=0 ?

The solutions to the equation x^2+(-x+3)^2-4x-21=0 are x=6,x=-1

Find the zeros of x^2+(-x+3)^2-4x-21=0

The zeros of x^2+(-x+3)^2-4x-21=0 are x=6,x=-1

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x^{2} + (- x + 3)^{2} - 4 x - 21 = 0

x = 6, x = - 1

Solution steps

x^{2} + (- x + 3)^{2} - 4 x - 21 = 0

Expand

x^{2} + (- x + 3)^{2} - 4 x - 21 : 2 x^{2} - 10 x - 12

2 x^{2} - 10 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 2 ( - 12 )}{2 \cdot 2}

(- 10)^{2} - 4 \cdot 2 (- 12) = 14

x_{1, 2} = \frac{- ( - 10 ) \pm 14}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 14}{2 \cdot 2}, x_{2} = \frac{- ( - 10 ) - 14}{2 \cdot 2}

x = \frac{- ( - 10 ) + 14}{2 \cdot 2} : 6

x = \frac{- ( - 10 ) - 14}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = 6, x = - 1

2 x^{2} - 9 x + 7

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

2 p^{2} = 8

2 x^{2} = - 5

2 x^{2} = 38

What are the solutions to the equation x^2+(-x+3)^2-4x-21=0 ?
The solutions to the equation x^2+(-x+3)^2-4x-21=0 are x=6,x=-1
Find the zeros of x^2+(-x+3)^2-4x-21=0
The zeros of x^2+(-x+3)^2-4x-21=0 are x=6,x=-1