Answer:
Step-by-step explanation:
To find how many real solutions it has you have to find the solutions.
The quadratic formula is x=[tex]\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
(the A with a weird line over it shouldn't be there but i cant get rid of it)
for the equation you gave me, you subtract 5 from both sides so it is equal to 0. It should be:
[tex]2x^{2} -4x-2[/tex]
This equation is now in the format:
[tex]ax^{2} +bx+c[/tex]
Now plug in what you have for a, b, and c (a= 2, b= -4, c= -2)
This will give you:
[tex]x=\frac{-(-4)±\sqrt{(-4)^{2}-4(2)(-2) } } {2(2)}[/tex]
(again ignore that letter)
Now simplify it to:
[tex]x=\frac{4±\sqrt{(16-(-16) } } {4}[/tex]
[tex]x=\frac{4±\sqrt{32} } {4}[/tex]
[tex]x=\frac{4±4\sqrt{2} } {4}[/tex]
[tex]x=1±\sqrt{2}[/tex]
There are 2 real solutions because 1 plus the square root of 2 and 1 minus the square root of two are both real. (If it was 1 plus or minus the square root of a negative number it would be 0 real solutions because you cant take the square root of a negative number)
Sorry if this is confusing
