What are the solutions to the equation -a^2-2a+3=0 ?

The solutions to the equation -a^2-2a+3=0 are a=-3,a=1

Find the zeros of -a^2-2a+3=0

The zeros of -a^2-2a+3=0 are a=-3,a=1

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- a^{2} - 2 a + 3 = 0

a = - 3, a = 1

Solution steps

- a^{2} - 2 a + 3 = 0

Solve with the quadratic formula

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

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

a_{1, 2} = \frac{- ( - 2 ) \pm 4}{2 ( - 1 )}

Separate the solutions

a_{1} = \frac{- ( - 2 ) + 4}{2 ( - 1 )}, a_{2} = \frac{- ( - 2 ) - 4}{2 ( - 1 )}

a = \frac{- ( - 2 ) + 4}{2 ( - 1 )} : - 3

a = \frac{- ( - 2 ) - 4}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

a = - 3, a = 1

a^{2} + 10 a - 12 = 0

a = a^{2} + (2 - \frac{4}{3} a)^{2} (\frac{3}{4})

(5 - x) (- 2 - x) + 12 = 0

5 x^{2} + 12 = 0

x^{2} - 16 x + 23 = 0

What are the solutions to the equation -a^2-2a+3=0 ?
The solutions to the equation -a^2-2a+3=0 are a=-3,a=1
Find the zeros of -a^2-2a+3=0
The zeros of -a^2-2a+3=0 are a=-3,a=1