What are the solutions to the equation 4r^2+8r+4=0 ?

The solutions to the equation 4r^2+8r+4=0 are r=-1

Find the zeros of 4r^2+8r+4=0

The zeros of 4r^2+8r+4=0 are r=-1

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4 r^{2} + 8 r + 4 = 0

r = - 1

Solution steps

4 r^{2} + 8 r + 4 = 0

Solve with the quadratic formula

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

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

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

r = \frac{- 8}{2 \cdot 4}

\frac{- 8}{2 \cdot 4} = - 1

r = - 1

The solution to the quadratic equation is:

r = - 1

(6 X - 4) (3 X + 5) = (4 X - 9) (3 X + 5)

m^{2} - 6 m + 8 = 0

4 x (x + 4) = 5 (x - 9)

9 x^{2} - 5 x - 2 = - 1 - x

0 = x^{2} - x - 42

What are the solutions to the equation 4r^2+8r+4=0 ?
The solutions to the equation 4r^2+8r+4=0 are r=-1
Find the zeros of 4r^2+8r+4=0
The zeros of 4r^2+8r+4=0 are r=-1