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

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

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

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

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

a = - 2, a = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

a_{1} = \frac{- ( - 1 ) + 3}{2 ( - 1 )}, a_{2} = \frac{- ( - 1 ) - 3}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

a = - 2, a = 1

\frac{z}{5} = \frac{3}{10} z + 7

t^{2} - 6 < - t + 14

- u + 15 \geq 9 (- u - 7) - 2

\frac{a}{- 16} + 101 = 99

∣ x + 9 ∣ \geq 2

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