Tu ejercicio es:

3w^2 + 17w + 10 = 0

La fórmula general para ecuaciones cuadráticas es:

x = [-b ± √(b^2 - 4ac)] / 2a

donde:

a = 3

b = 17

c = 10

Resolviendo queda:

x = [-17 ± √(17^2 - 4 * 3 * 10)] / (2 * 3)

x = [-17 ± √(289 - 120) / 6

x = [-17 ± √169] / 6

x = [-17 ± 13] / 6

Raíces:

x1 = [-17 + 13] / 6 = -4/6 = -2/3

x2 = [-17 - 13] / 6 = -30/6 = -5

https://www.youtube.com/watch?v=Y5zCUpQy6Rw

First method: by completing square

3w² + 17w + 10 = 0

3.[w² + (17/3).w + (10/3)] = 0

w² + (17/3).w + (10/3) = 0

w² + (17/3).w + (17/6)² - (17/6)² + (10/3) = 0

w² + (17/3).w + (17/6)² - (289/36) + (120/36) = 0

w² + (17/3).w + (17/6)² - (169/36) = 0

w² + (17/3).w + (17/6)² = 169/36

[w + (17/6)]² = (± 13/6)²

w + (17/6) = ± (13/6)

w = - (17/6) ± (13/6)

w = (- 17 ± 13)/6

w₁ = (- 17 + 13)/6 → w₁ = - 4/6 → w₁ = - 2/3

w₂ = (- 17 - 13)/6 → w₂ = - 30/6 → w₂ = - 5

Second method: by combination

3w² + 17w + 10 = 0 → you can see that: (3 * 10) = 30 → and you know that: 30 = 15 * 2

3w² + (15w + 2w) + 10 = 0

3w² + 15w + 2w + 10 = 0

(3w² + 15w) + (2w + 10) = 0

3w.(w + 5) + 2.(w + 5) = 0

(3w + 2).(w + 5) = 0

First case: (3w + 2) = 0 → 3w + 2 = 0 → 3w = - 2 → w = - 2/3

Second case: (w + 5) = 0 → w + 5 = 0 → w = - 5

Third method: with the discriminant

3w² + 17w + 10 = 0

Polynomial like: ax² + bx + c, where:

a = 3

b = 17

c = 10

Δ = b² - 4ac (discriminant)

Δ = 17² - (4 * 3 * 10)

Δ = 289 - 120 = 169 = 13²

w₁ = (- b - √Δ)/2a = (- 17 - 13)/(2 * 3) = - 30/6 = - 5

w₂ = (- b + √Δ)/2a = (- 17 + 13)/(2 * 3) = - 4/6 = - 2/3