(2x^2+7x-8)(2x^2+7x-3)-6=0

Question

(2x^2+7x-8)(2x^2+7x-3)-6=0

Answer 1

2x^2+7x=t
(t-8)(t-3)-6=0
t²-11t+24-6=0
t²-11t+18=0
D=121-4×18=121-72=49 √D=7
t1=(11-7)/2=2
2x²+7x-2=0
D=49+16=65 √D=√65
x1=(-7+√65)/4
x2=(-7-√65)/4
t2=18/2=9
2x²+7x-9=0
D=49+18×4=121 √D=11
x1=(-7+11)/4=1
x2=(-18)/4