What are the solutions to the equation 2r^2+8r+6=0 ?

The solutions to the equation 2r^2+8r+6=0 are r=-1,r=-3

Find the zeros of 2r^2+8r+6=0

The zeros of 2r^2+8r+6=0 are r=-1,r=-3

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2 r^{2} + 8 r + 6 = 0

r = - 1, r = - 3

Solution steps

2 r^{2} + 8 r + 6 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 6 = 4

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

Separate the solutions

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

r = \frac{- 8 + 4}{2 \cdot 2} : - 1

r = \frac{- 8 - 4}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

r = - 1, r = - 3

x (x - 2) = 3

(x) = 0.3 x^{2} - 144 x + 30178

3 x^{2} + 7 x = 4

x^{2} = 4 x

(3 n - 1)^{2} = 64

What are the solutions to the equation 2r^2+8r+6=0 ?
The solutions to the equation 2r^2+8r+6=0 are r=-1,r=-3
Find the zeros of 2r^2+8r+6=0
The zeros of 2r^2+8r+6=0 are r=-1,r=-3