Question

2x^2 + x – 1 = 9 
Does it have any real solutions? If so what are they?

Answer 1

2x^2 + x – 1 = 9
2x^2+x-1-9=0
2x^2+x-10=0
a=2
b=1
c=-10

Delta=b^2-4ac
Delta=1-4*2*(-10)
delta=1+80
delta=81
rad 81=9

x1=-b-rad 81 /2a
x1=-1-9/2*2
x1=-10/4
x1=-5/2

x2=-b+rad 81 /2a
x2=-1+9/2*2
x2=8/4
x2=2

Answer 2

`2x^2 + x - 1 = 9 \\ \\2x^2 + x -1 - 9=0\\ \\2x^2 + x-5x +5x -10 = 0 \\ \\ 2x^2 -4x +5x -10 = 0\\ \\ 2x(x -2) +5(x -2) = 0\\ \\(x-2)(2x+5)=0\\ \\x-2 =0 \ \ or \ \ 2x+5 =0 \\ \\x=2 \ \ or \ \ 2x =-5 \\ \\x=2 \ \ or \ \ x =-(5)/(2) \\ \\Answer : \ x=2 \ \ or \ \ x=-(5)/(2)`