how do you solve (4-3i)times(-5+2i) and also solve the equation y=-3ysquared+14)?

Answer 1

Question 1
(4 - 3i)(- 5 + 2i)
- 20 + 8 i + 15 i - 6i²
- 20 + 23 i + 6
- 14 + 23 i

Question 2
Is this
y = - 3y² + 14?

3y² + y - 14 = 0
y = [- 1 ± √(1 + 168) ] / 6
y = [- 1 ± √(169) ] / 6
y = 12/6 ,y = - 14/6
y = 2 , y = - 7/3

Answer 2

(4 - 3i) * (-5 + 2i)
= -20 +15i + 8i - 6i^2
= -20 +23i + 6
= -14 + 23i

y = - 3y^2 + 14
=> 3y^2 + y - 14 = 0
=> 3y^2 - 6y + 7y - 14 = 0
=> 3y(y - 2) + 7(y - 2) =0
=> (3y + 7) ( y - 2) = 0
=> 3y + 7 = 0 or y - 2 = 0
=> y = - 7/3 or y = 2.

Answer 3

FOIL it. (4-3i)(-5-2i)= -20 -8i+15i+6i^2=-20+7i-6 since i^2=-1
so 6(-1)= -6

y= -3y^2+14
3y^2+1y-14=0
(3y + 7)(y - 2 )=0
So, 3y+7=0 or y-2=0
3y = -7 or y=2
y = -7/3 or y=2

Answer 4

-20 + 8i +15i - 6i^2
-6i^2 + 23i - 20



y = -3y^2 + 14
3y^2 + y - 14 = 0
(3y + 7)(y-2) = 0
3y + 7 = 0 and y - 2 = 0
3y = -7 and y = 2
y = -7/3 and y = 2