X² - 15 x + 36 = 0
x² - 12x - 3x + 36 = 0
x( x - 12) - 3( x - 12) = 0
(x - 3)( x - 12) = 0
x - 3 = 0
x = 3
or
x - 12 = 0
x = 12
The solutions are 3 and 12
Answer:
2ab(7a^4b^2+ a^5×b-2a)
start with multiple 2ab into the equation
(2ab×7a^4b^2)+(2ab×a^5×b)-(2ab×2a)
((2×7)a^(4+1)b^(2+1))+(2×a^(5+1)×b^(1+1))-((2×2)a^(1+1)×b)
14×a^5×b^3+2×a^6×b^2-4×a^2×b
Answer:
3i
Step-by-step explanation:
Using the rule of radicals
×
⇔ 
and
= i
Given

= 
=
×
× 
= 3 ×
× i
= 3i
Here you go you little troll <span>191630195140</span>