What are the solutions to the equation (-a)^2-4(1)(-2a)=0 ?

The solutions to the equation (-a)^2-4(1)(-2a)=0 are a=0,a=-8

Find the zeros of (-a)^2-4(1)(-2a)=0

The zeros of (-a)^2-4(1)(-2a)=0 are a=0,a=-8

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(- a)^{2} - 4 (1) (- 2 a) = 0

a = 0, a = - 8

Solution steps

(- a)^{2} - 4 (1) (- 2 a) = 0

Expand

(- a)^{2} - 4 (1) (- 2 a) : a^{2} + 8 a

a^{2} + 8 a = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 0 = 8

a_{1, 2} = \frac{- 8 \pm 8}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- 8 + 8}{2 \cdot 1}, a_{2} = \frac{- 8 - 8}{2 \cdot 1}

a = \frac{- 8 + 8}{2 \cdot 1} : 0

a = \frac{- 8 - 8}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

a = 0, a = - 8

v, v^{2} = u^{2} + 2 a s

x, (x - 0)^{2} = 4 p (y - 10)

2 y^{2} - y - 7 = 0

x, f (x + 1) = - x^{2} + x + 1

x, x^{2} - 3 x^{2} y = C

What are the solutions to the equation (-a)^2-4(1)(-2a)=0 ?
The solutions to the equation (-a)^2-4(1)(-2a)=0 are a=0,a=-8
Find the zeros of (-a)^2-4(1)(-2a)=0
The zeros of (-a)^2-4(1)(-2a)=0 are a=0,a=-8