Factor 3y2 + 29y − 10. 3y2 + 29y – 10 = (________ − 1)( _________ + 10)I want to see the work too because i am a bit confused.

quadratic trinomial can be represent as product
ax² + bx + c =a(x - x1)(x - x2), where x1 and x2 are roots of quadratic equation ax² + bx + c=0,
so to solve this problem we need to find roots of 
3y²+29y-10=0

y=(-b+/-√(b²-4ac))/2a
y=(-29+/-√(29²-4*3*(-10)))/(2*3)
y=(-29+/-√(961)/(6)
y=(-29+/-31/(6)
y1=-60/6=-10
y2=2/6=1/3

3(y-1/3)(y+10)=(3y-1)(y+10)
check:   (3y-1)(y+10)=3y²-y+30y-10=3y²+29y-10