What are the solutions to the equation 2r^2+8r+8=0 ?

The solutions to the equation 2r^2+8r+8=0 are r=-2

Find the zeros of 2r^2+8r+8=0

The zeros of 2r^2+8r+8=0 are r=-2

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2 r^{2} + 8 r + 8 = 0

r = - 2

Solution steps

2 r^{2} + 8 r + 8 = 0

Solve with the quadratic formula

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

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

r_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 2}

r = \frac{- 8}{2 \cdot 2}

\frac{- 8}{2 \cdot 2} = - 2

r = - 2

The solution to the quadratic equation is:

r = - 2

2 x^{2} - 6 x + 6 = c^{1} (x^{2} + x - 1) + c^{2} (x^{2} - x + 1)

(2 x - 10) (3 x + 12) = 0

a^{2} - 2 a = 2

x^{2} + 4 x - 16

x^{2} + 2 x - 40 = 0

What are the solutions to the equation 2r^2+8r+8=0 ?
The solutions to the equation 2r^2+8r+8=0 are r=-2
Find the zeros of 2r^2+8r+8=0
The zeros of 2r^2+8r+8=0 are r=-2