При каких значения а произведение корней уравнения x^2+5ax+a^2+4a+10=0 равно 10?

Объяснение:

a²+4a+10=10

a(a+4)=0

при a=0 либо при a= -4

4х²=2х-3
4х²-2х+3=0         
D=(-2)²-4×4×3=4-48=-44        D<0, уравнение не имеет корней
----------------------------------------------------------------------------
5х²+26х=24
5х²+26х-24=0
D=26²-4×5×(-24)=676+480=1156          D>0
х₁=
х₂=
х₁=0,8
х₂=-6
-------------------------------------------------------------------------
3х²-5х=0
D=5²-4×3×0=25-0=25          D>0
х₁=
х₂=
х₁=1,667
х₂=0
--------------------------------------------------------------------
6-2х²=0
-2х²+6=0
D=0²-4×(-2)×6=0+48=48        D>0
х₁=
х₂=
х₁=-1,732
х₂=1,732
------------------------------------------------------------------
t²=35-2t
t²+2t-35=0
D=2²-4×1×(-35)=4+140=144
t₁=
t₂=
t₁=5
t₂=-7

10-4(2x+5)=6y-13

4y-63=5(4x-2y)+2

 

4y-63=5(4x-2y)+2

4y-63-20x+10y-2=0

14y-65-20x=0

x=0.7y-3.25

 

10-4(2x+5)=6y-13

3-8x-6y=0

3-8*(0.7y-3.25)-6y=0

3-5.6y+26-6y=0

29-11.6y=0

y=29/11.6
y=2.пятых

 

14y-65-20x=0

14*2.5-65-20x=0

-30-20x=0

x=-30/20

x=-1.5

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