What are the solutions to the equation (3-x)(-3-x)+10=0 ?

The solutions to the equation (3-x)(-3-x)+10=0 are x=i,x=-i

Find the zeros of (3-x)(-3-x)+10=0

The zeros of (3-x)(-3-x)+10=0 are x=i,x=-i

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(3 - x) (- 3 - x) + 10 = 0

x = i, x = - i

Solution steps

(3 - x) (- 3 - x) + 10 = 0

Expand

(3 - x) (- 3 - x) + 10 : x^{2} + 1

x^{2} + 1 = 0

Solve with the quadratic formula

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

Simplify

0^{2} - 4 \cdot 1 \cdot 1 : 2 i

x_{1, 2} = \frac{- 0 \pm 2 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 2 i}{2 \cdot 1}, x_{2} = \frac{- 0 - 2 i}{2 \cdot 1}

x = \frac{- 0 + 2 i}{2 \cdot 1} : i

x = \frac{- 0 - 2 i}{2 \cdot 1} : - i

The solutions to the quadratic equation are:

x = i, x = - i

12 x^{2} + 48 x + 45 = 0

4 x^{2} - 6 x - 8 = - 2 x - 5

4 x^{2} - 11 x - 45 = 0

15495.87 x^{2} + 231829.74 x - 19620 = 0

x, x^{2} + y^{2} + a x + b y + c = 0

What are the solutions to the equation (3-x)(-3-x)+10=0 ?
The solutions to the equation (3-x)(-3-x)+10=0 are x=i,x=-i
Find the zeros of (3-x)(-3-x)+10=0
The zeros of (3-x)(-3-x)+10=0 are x=i,x=-i