Question

9y^2-6y+1=0 -3x^2-2x+5=0 5x^2-8x+3=0 хелп

Answer 1

9у²-6у+1=0
D=36-36=0
y=1/3
-3x²-2x+5=0
D=4+60=64
x1=-5/3
x2=1
5x²-8x+3=0
D=64-60=4
х1=-1
х2=-3/5

Answer 2

It looks like you have provided three different quadratic equations, each in the form of $ax^2 + bx + c = 0$, and you're looking for help with solving them. I'll go through each equation step by step.

Equation 1: $9y^2 - 6y + 1 = 0$ This is a quadratic equation in terms of $y$. You can solve it using the quadratic formula:

The quadratic formula is given by: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For your equation, $a = 9$, $b = -6$, and $c = 1$. Plugging these values into the formula:

$y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9}$ $y = \frac{6 \pm \sqrt{36 - 36}}{18}$ $y = \frac{6 \pm 0}{18}$ $y = \frac{6}{18}$ $y = \frac{1}{3}$

So, the solution for Equation 1 is $y = \frac{1}{3}$.

Equation 2: $-3x^2 - 2x + 5 = 0$ This is also a quadratic equation. You can use the quadratic formula again:

For this equation, $a = -3$, $b = -2$, and $c = 5$. Plugging these values into the formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot (-3) \cdot 5}}{2 \cdot (-3)}$ $x = \frac{2 \pm \sqrt{4 + 60}}{-6}$ $x = \frac{2 \pm \sqrt{64}}{-6}$ $x = \frac{2 \pm 8}{-6}$

So, the two solutions for Equation 2 are: $x = -\frac{5}{3}$ (using the plus sign) $x = 2$ (using the minus sign)

Equation 3: $5x^2 - 8x + 3 = 0$ Once again, this is a quadratic equation. Applying the quadratic formula:

For this equation, $a = 5$, $b = -8$, and $c = 3$. Plugging these values into the formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 5 \cdot 3}}{2 \cdot 5}$ $x = \frac{8 \pm \sqrt{64 - 60}}{10}$ $x = \frac{8 \pm \sqrt{4}}{10}$ $x = \frac{8 \pm 2}{10}$

So, the two solutions for Equation 3 are: $x = \frac{1}{5}$ (using the plus sign) $x = \frac{3}{5}$ (using the minus sign)

I've solved all three quadratic equations for you. If you have any more questions or need further assistance, feel free to ask!