Question

4х2+5х-6=0х2+15х+44=0?????​

Answer 1

Ответ:

1) x₁=(-2)             x₂ =0,75

2) x₁=-11          x₂=-4

Объяснение:

1) 4х²+5х-6=0

D=5²-4*(-6)*4=25+16*6=25+96=121=11²

$x_{1,2}=\frac{-5\pm11}{2*4}$    

$x_1=\frac{-5-11}{8}$    $x_2=\frac{6}{8}$

$x_1=-\frac{16}{8}$       $x_2=\frac{3}{4}$

x₁=(-2)             x₂ =0,75

2) х²+15х+44=0

D=15²-4*44=225-176=49=7²

$x_{1,2}=\frac{-15\pm7}{2}$

$x_1=\frac{-22}{2}$    $x_2=\frac{-15+7}{2}$

x₁=-11          x₂=-8:2

x₁=-11          x₂=-4

Answer 2

It seems like you have two equations here, and you're asking about them:

1. 4x^2 + 5x - 6 = 0
2. 0x^2 + 15x + 44 = 0

Let's solve each equation separately:

1. 4x^2 + 5x - 6 = 0

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

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 4, b = 5, and c = -6. Plugging these values into the formula:

x = (-5 ± √(5² - 4 * 4 * -6)) / (2 * 4)

x = (-5 ± √(25 + 96)) / 8

x = (-5 ± √121) / 8

x = (-5 ± 11) / 8

So, you have two solutions:

x1 = (11 - 5) / 8 = 6/8 = 3/4 x2 = (-5 - 11) / 8 = -16/8 = -2

2. 0x^2 + 15x + 44 = 0

This equation simplifies to:

15x + 44 = 0

Now, subtract 44 from both sides:

15x = -44

Finally, divide by 15 to isolate x:

x = -44/15

So, the solutions to the second equation are:

x = -44/15

In summary:

1. The solutions to the first equation are x = 3/4 and x = -2.
2. The solution to the second equation is x = -44/15.