Hi, can anyone please help with this? I'm having a very hard time with this specific problem plus math is my least strong subject and it would mean so much if someone could help. Thank you!!

Step-by-step explanation:

For reference the quadratic equations looks like this

(-b±)/2a

Amy's mistake (and solution)

We can see that when amy mulitplied c she wrote 20 instead of -20 (in step 1). She can fix this by writing (4*1*-20) instead

Richard's mistake (and solution)

When richard messed up on step 5 when he wrote (6-2i)/2. You only would write the i if the negative was in the radical (but in this case it's outside). You would fix this by just getting rid of the i entirely (and keeping the 2)

I'm not really too sure how you would do three but I'll give it a shot anyways

our equation will look like this: x²+2x

which means our quadratic equation would look like this

When I write +/- I mean plus or minus giving us -2 and 0 as our solutions

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.