Question

R a)-2x- 15 0, B)x+10x +9 0, a) 3x2+14x + 16 0

Answer 1

Ответ:

а) -2х=15

х=15÷(-2)

х=-7.5

В) 11х=-9

х=-9/11

а)D=14^2-4×3×16=196-192=4=2^2

x12= -14+-2/6

x1=-2

x2=-8/3=-2целых2/3

Объяснение:

Все есть удачи Если что обращайся помогу чем смогу СДЕЛАЙ ЛУЧШИМ

Answer 2

It looks like you've provided three quadratic equations and asked for something related to them. However, there seems to be a bit of confusion in how the equations are written. Let's clarify and solve each equation:

1. Equation A: -2x - 15 = 0

   To solve for x, add 15 to both sides and then divide by -2:

   $-2x - 15 + 15 = 0 + 15$ $-2x = 15$ $x = \frac{15}{-2}$ $x = -\frac{15}{2}$

   So, the solution for Equation A is $x = -\frac{15}{2}$.

2. Equation B: x + 10x + 9 = 0

   Combine like terms and solve for x:

   $11x + 9 = 0$ $11x = -9$ $x = \frac{-9}{11}$

   The solution for Equation B is $x = \frac{-9}{11}$.

3. Equation C: 3x^2 + 14x + 16 = 0

   To solve for x in a quadratic equation, you can use the quadratic formula:

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

   For the given equation $3x^2 + 14x + 16 = 0$, the coefficients are $a = 3$, $b = 14$, and $c = 16$.

   $x = \frac{-14 \pm \sqrt{14^2 - 4(3)(16)}}{2(3)}$ $x = \frac{-14 \pm \sqrt{196 - 192}}{6}$ $x = \frac{-14 \pm \sqrt{4}}{6}$ $x = \frac{-14 \pm 2}{6}$

   There are two solutions: $x_1 = \frac{-14 + 2}{6} = \frac{-12}{6} = -2$

   $x_2 = \frac{-14 - 2}{6} = \frac{-16}{6} = -\frac{8}{3}$

   So, the solutions for Equation C are $x = -2$ and $x = -\frac{8}{3}$.

Please double-check the equations if there was any mistake in transcription. If you have further questions or need clarification, feel free to ask!