Question

2 ( х+3) в 2 - 5 ( х+3)+ 2= 0

Answer 1

2(x+3)^2-5(x+3)+2=0
2(x^2+6x+9)-5x-15=0
2x^2+12x+18-5x-15=0
2x^2-5x+3=0 |:2
x^2-2,5x+1,5=0
x1+x2=2,5
x1*x2=1,5
x1=1
x2=1,5

Answer 2

I can help you with your math problem. The equation you wrote is equivalent to:

$$2(x+3)^2-5(x+3)^2=0$$

To solve this equation, we can use the following steps:

- Simplify the equation by expanding the brackets and combining like terms:

$$2(x^2+6x+9)-5(x^2+6x+9)=0$$ $$-3x^2-18x-18=0$$

- Divide both sides by -3 to get a simpler equation:

$$x^2+6x+6=0$$

- Use the quadratic formula to find the values of x that satisfy the equation:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

- Substitute the values of a, b, and c from the equation:

$$x=\frac{-6\pm\sqrt{6^2-4(1)(6)}}{2(1)}$$

- Simplify the expression and get the final answer:

$$x=\frac{-6\pm\sqrt{12}}{2}$$ $$x=-3\pm\sqrt{3}$$

Therefore, the equation has two solutions: $$x=-3+\sqrt{3}$$ and $$x=-3-\sqrt{3}$$.

I hope this helps you understand how to solve the equation. If you have any other questions, feel free to ask me.Ответь+подробно.+2+(+х+3)+в+2+-+5+(+х+3)++2=+0