What are the solutions to the equation 4r^2+15r+7=-1 ?

The solutions to the equation 4r^2+15r+7=-1 are r=(-15+sqrt(97))/8 ,r=(-15-sqrt(97))/8

Find the zeros of 4r^2+15r+7=-1

The zeros of 4r^2+15r+7=-1 are r=(-15+sqrt(97))/8 ,r=(-15-sqrt(97))/8

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4 r^{2} + 15 r + 7 = - 1

r = \frac{- 15 + 97}{8}, r = \frac{- 15 - 97}{8}

Solution steps

4 r^{2} + 15 r + 7 = - 1

Move

1

to the left side

4 r^{2} + 15 r + 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 4 \cdot 8}{2 \cdot 4}

1 5^{2} - 4 \cdot 4 \cdot 8 = 97

r_{1, 2} = \frac{- 15 \pm 97}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- 15 + 97}{2 \cdot 4}, r_{2} = \frac{- 15 - 97}{2 \cdot 4}

r = \frac{- 15 + 97}{2 \cdot 4} : \frac{- 15 + 97}{8}

r = \frac{- 15 - 97}{2 \cdot 4} : \frac{- 15 - 97}{8}

The solutions to the quadratic equation are:

r = \frac{- 15 + 97}{8}, r = \frac{- 15 - 97}{8}

3 x^{2} + 2 (x^{2} - 20) = 16 - 2 x^{2}

r, a = π r^{2}

x^{2} - 6 x - 20 = 0

2 x^{2} + 13 x - 73 = x^{2} - 19

x, z^{3} - 3 x^{2} y + 6 x yz = 0

What are the solutions to the equation 4r^2+15r+7=-1 ?
The solutions to the equation 4r^2+15r+7=-1 are r=(-15+sqrt(97))/8 ,r=(-15-sqrt(97))/8
Find the zeros of 4r^2+15r+7=-1
The zeros of 4r^2+15r+7=-1 are r=(-15+sqrt(97))/8 ,r=(-15-sqrt(97))/8