What substitution should be used to rewrite 6 (x+5) ^2+5 (x+5) - 4=0 as a quadratic equation

Step-by-step explanation:

6 (x+5) ^2 + 5 (x+5) - 4=0

let x+5 = y

6y^2 + 5y - 4=0

6y^2 + 8y - 3y-4 = 0

2y (3y+4) - 1 (3y+4) = 0

(2y-1) (3y+4) = 0

2y-1=0 or 3y+4=0

y=1/3 or - 4/3

Recall that x+5=y

When y=1/3

x+5=1/3

x = - 14/3

when y = - 4/3

x+5=-4/3

x = - 11/3

u = (x+5)