Question

0.5у×(2у-1)-(0.5-2у)+1=0

Answer 1

$0.5y(2y-1)-(0.5-2y)+1=0$
$y^{2} -0.5y-0.5+2y+1=0$
$y^{2} +1.5y+0.5=0$
$2y^{2} +3y+1=0$
$D=9-8=1$
$y1= \frac{-3+1}{4} =-0.5$

Answer 2

Let's solve the given equation step by step:

0.5у × (2у - 1) - (0.5 - 2у) + 1 = 0

First, let's simplify the expression within the parentheses:

0.5у × 2у - 0.5у - (0.5 - 2у) + 1 = 0

Now, let's simplify further:

у² - 0.5у - 0.5 + 2у + 1 = 0

у² + 1.5у + 0.5 = 0

To solve this quadratic equation, we can use the quadratic formula:

у = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 1, b = 1.5, and c = 0.5.

Plugging these values into the quadratic formula:

у = (-(1.5) ± √((1.5)² - 4(1)(0.5))) / (2(1))

Simplifying further:

у = (-1.5 ± √(2.25 - 2)) / 2

у = (-1.5 ± √(0.25)) / 2

Now, let's calculate the square root:

у = (-1.5 ± 0.5) / 2

This gives us two possible solutions:

Solution 1: у = (-1.5 + 0.5) / 2 = -1 / 2 = -0.5

Solution 2: у = (-1.5 - 0.5) / 2 = -2 / 2 = -1

Therefore, the solutions to the equation 0.5у × (2у - 1) - (0.5 - 2у) + 1 = 0 are у = -0.5 and у = -1.