Both A and C would be solutions to the equation.
In order to solve for this you must first get the equation equal to 0.
2x^2+5x+8=6 ----> subtract 6 from both sides
2x^2 + 5x + 2 = 0
Now knowing this we can use the coefficients of each one in descending order of power as a, b and c.
a = 2 (because it is the coefficient to x^2)
b = 5 (because it is the coefficient to x)
c = 2 (because it is the end number)
Now we can plug these values into the quadratic equation.





or -1/2 for the first answer

or -2 for the second answer
Answer:
im taking the same test lol
Step-by-step explanation:
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Hope it helps, despite being this long!
<span>f(x) = 5x+1
f(-4) = 5(-4) + 1 = -20 + 1 = -19
hope that helps
</span>