Solve the following quadratic inequality 2x^2+3x-5>0
2 answers:
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I hope this helps you
The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=
x^2+7x+18
a=1
b=7
c=18
x=
x=
x=
i=√-1
x=
the zerose would be
x=
or 
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=
x^2+7x+18
a=1
b=7
c=18
x=
x=
x=
i=√-1
x=
the zerose would be
x=