Question

2(х^2-40)=-х^2+6(х+4)+1

Answer 1

2(х²-40)=-х²+6(х+4)+1
2х²-80=-х²+6х+24+1
2х²-80+х²-6х-24-1=0
3х²-6х-105=0
Д=а²-4ав=36-4*3*(-105)=1296

х1= 6+36/6=7
х2=6-36/6=-5
/-дробь

Answer 2

Let's simplify and solve the given equation step by step:

2(x^2 - 40) = -x^2 + 6(x + 4) + 1

First, we'll distribute the multiplication on both sides:

2x^2 - 80 = -x^2 + 6x + 24 + 1

Next, we'll simplify by combining like terms:

3x^2 - 6x - 55 = 0

Now we have a quadratic equation in standard form (ax^2 + bx + c = 0), where a = 3, b = -6, and c = -55. We can solve for x using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Substituting the values, we get:

x = (-(-6) ± sqrt((-6)^2 - 4(3)(-55))) / 2(3) x = (6 ± sqrt(576)) / 6 x = (6 ± 24) / 6

So, x can have two possible values:

x = 5, or x = -5/3

Therefore, the solution set for the given equation is {5, -5/3}.