You can put this solution on YOUR website! 36z^2+96z+15=0
divide each side by 3
12z^2+32z+5=0
The product of the outer coefficients is 12*5 = 60
We need a pair of numbers whose product is 60 and sum is 32.
The numbers 2 and 30 work.
so we can rewrite 12z^2+32z+5=0 as
12z^2+30z+2z+5=0
Now group 12z^2+30z and +2z+5
6z(2z+5) + 2z+5
6z(2z+5) + 1(2z+5)
(6z+1)(2z+5)
First solve 6z+1 = 0
6z = -1
z = -1/6
Second solve 2z+5 = 0
2z = -5
z = -5/2
