Question

A) x² + 3x – 130 = 0; ​

Answer 1

Решение:

x² + 3x – 130 = 0

а=1, b=3, c=-130

D=b²-4ac

D=3²-4*1*(-130)=9+520=529=23²

x₁=(-b+√D)/(2a)=(-3+23)/(2*1)=20/2=10

х₂=(-b-√D)/(2a)=(-3-23)/(2*1)=-26/2=-13

Ответ: 10; -13.

Answer 2

Ответ:  x² + 3x – 130 = 0

D = 9 + 520 = 529

x = (-3 + 23) : 2 = 10

x = (-3 - 23) : 2 = - 13

Объяснение:

Answer 3

To solve the quadratic equation $x^2 + 3x - 130 = 0$, we can use the quadratic formula:

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

Here, $a$, $b$, and $c$ are the coefficients of the quadratic equation. In this case, $a = 1$, $b = 3$, and $c = -130$. Substituting these values into the formula, we have:

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-130)}}{2(1)}$

Simplifying further:

$x = \frac{-3 \pm \sqrt{9 + 520}}{2}$ $x = \frac{-3 \pm \sqrt{529}}{2}$ $x = \frac{-3 \pm 23}{2}$

This gives us two possible solutions for $x$:

$x_1 = \frac{-3 + 23}{2} = \frac{20}{2} = 10$ $x_2 = \frac{-3 - 23}{2} = \frac{-26}{2} = -13$

Therefore, the solutions to the equation $x^2 + 3x - 130 = 0$ are $x = 10$ and $x = -13$.