Question

Х в квадрате = -15х -56

Answer 1

х²+15х+56=0

Д=в²-4*а*с=15²-4*1*56=1

х=-15±1/2

х=-15+1/2=-7

х=-15-1/2=-8

Ответ х1=-8,х2=-7

Answer 2

To solve the quadratic equation $x^2 = -15x - 56$, we need to rewrite it in standard quadratic form, which is $ax^2 + bx + c = 0$. Then, we can use the quadratic formula to find the values of $x$.

Given the equation $x^2 = -15x - 56$, let's move all terms to the left-hand side:

$x^2 + 15x + 56 = 0$

Now, we can use the quadratic formula to find the solutions for $x$:

The quadratic formula states that for an equation in the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

In our case, $a = 1$, $b = 15$, and $c = 56$. Plugging these values into the formula, we get:

$x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 1 \cdot 56}}{2 \cdot 1}$

Simplifying further:

$x = \frac{-15 \pm \sqrt{225 - 224}}{2}$

$x = \frac{-15 \pm \sqrt{1}}{2}$

Now, we have two possibilities for $x$:

1. When $x = \frac{-15 + \sqrt{1}}{2} = \frac{-15 + 1}{2} = \frac{-14}{2} = -7$.

2. When $x = \frac{-15 - \sqrt{1}}{2} = \frac{-15 - 1}{2} = \frac{-16}{2} = -8$.

Thus, the solutions to the equation are $x = -7$ and $x = -8$.