Answer :
Answer:
Step-by-step explanation:
Let's solve for
z:
=
−
20
±
2
0
2
−
4
⋅
1
⋅
(
625
2
+
400
−
36
)
2
⋅
1
z=
2⋅1
−20±
20
2
−4⋅1⋅(625y
2
+400y−36)
=
−
20
±
400
−
(
4
⋅
625
⋅
(
400
2
+
400
/
625
)
−
4
⋅
36
)
2
z=
2
−20±
400−(4⋅625⋅(400y
2
+400/625)−4⋅36)
=
−
20
±
400
−
(
10000
2
+
6400
−
144
)
2
z=
2
−20±
400−(10000y
2
+6400y−144)
=
−
20
±
−
10000
2
−
6400
+
544
2
z=
2
−20±
−10000y
2
−6400y+544
Since the discriminant
(
2
−
4
)
(b
2
−4ac) is negative, the solutions will involve imaginary numbers. Therefore, the solutions for
z will be complex numbers involving
y.
