What are the solutions to the equation 4r^2+10r+5=-r ?

The solutions to the equation 4r^2+10r+5=-r are r=(-11+sqrt(41))/8 ,r=(-11-sqrt(41))/8

Find the zeros of 4r^2+10r+5=-r

The zeros of 4r^2+10r+5=-r are r=(-11+sqrt(41))/8 ,r=(-11-sqrt(41))/8

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4 r^{2} + 10 r + 5 = - r

r = \frac{- 11 + 41}{8}, r = \frac{- 11 - 41}{8}

Decimal Notation

r = - 0.57460 \dots, r = - 2.17539 \dots

Solution steps

4 r^{2} + 10 r + 5 = - r

Move

r

to the left side

4 r^{2} + 11 r + 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

1 1^{2} - 4 \cdot 4 \cdot 5 = 41

r_{1, 2} = \frac{- 11 \pm 41}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- 11 + 41}{2 \cdot 4}, r_{2} = \frac{- 11 - 41}{2 \cdot 4}

r = \frac{- 11 + 41}{2 \cdot 4} : \frac{- 11 + 41}{8}

r = \frac{- 11 - 41}{2 \cdot 4} : \frac{- 11 - 41}{8}

The solutions to the quadratic equation are:

r = \frac{- 11 + 41}{8}, r = \frac{- 11 - 41}{8}

2 t^{2} + 4 t - 6 = - 3 t

4 x - 2 x^{2} = 1

0.57 x^{2} + 0.41 x - 0.87 = 0

a^{3} - (a - 1)^{3} = 169

2 a^{2} - \frac{7}{2} a - 18 = 0

What are the solutions to the equation 4r^2+10r+5=-r ?
The solutions to the equation 4r^2+10r+5=-r are r=(-11+sqrt(41))/8 ,r=(-11-sqrt(41))/8
Find the zeros of 4r^2+10r+5=-r
The zeros of 4r^2+10r+5=-r are r=(-11+sqrt(41))/8 ,r=(-11-sqrt(41))/8