Answer:
[tex]x=\frac{5}{2}, x=-1[/tex]
Step-by-step explanation:
Hi there!
We are given the following quadratic equation:
2x²-3x-5=0
And we want to solve it.
There are multiple ways to solve it, but we can solve this by factoring.
One way to factor an equation like this if the leading coefficient a (the coefficient in front of x²) is greater than one, is to multiply the coefficients a and c (the coefficient without any variable on it) together, find out which numbers multiply to ac and equal b (the coefficient in front of x (without any exponent)), and then factor the equation as (ax+n)(ax+m)=0, then solve for x, where m and n are the numbers that multiply to ac and equal b
To start, multiply 2 and -5 together (a is 2 and c is -5):
-10
Now find out which numbers multiply to -10 but equal -3 (-3 is b):
Those two numbers are -5 and 2
Now factor the equation like this:
(2x-5)(2x+2)=0
Now solve for x:
2x-5=0
Add 5 to both sides
2x=5
Divide both sides by 2
x=[tex]\frac{5}{2}[/tex]
2x+2=0
Subtract 2 from both sides
2x=-2
Divide both sides by 2
x=-1
So the answers are [tex]x=\frac{5}{2}, x=-1[/tex]
Hope this helps!
