Hi,
f(x)=2x^2+4x+5 y=intercept Pt(0,5)
the vertex form of a parabola, 
where(h,k) is the vertex
f(x)=2[(x^2 +2x} +5
f(x)=2[(x+ 1)^2 -1} +5
f(x)=2(x+ 1) -2 + 5
f(x)=2(x+ 1)+3
vertex Pt(-1,3), parabola opens upward (a=2 and is > than zero)
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
The discriminant -24 is less than zero. That means that there are no solutions among real numbers.
If you are a student of advanced school algebra and are aware about imaginary numbers, read on.
In the field of imaginary numbers, the square root of -24 is + or -  .
The solution is 
Here's your graph:
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