Question

1)y²-5y-24=02)4x²+3x-10=0​

Answer 1

Ответ:

Изи

Объяснение:

Обеснение на фото извиняюсь за плохой почерк

Answer 2

It seems like you've provided two quadratic equations. To better assist you, I can help you solve these equations for their respective variables. Let's solve each equation:

1. $y^2 - 5y - 24 = 0$:

To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, let's factor the equation:

$y^2 - 5y - 24 = (y - 8)(y + 3) = 0$

Setting each factor equal to zero gives us two possible solutions for $y$:

$y - 8 = 0 \Rightarrow y = 8$

$y + 3 = 0 \Rightarrow y = -3$

So, the solutions for this equation are $y = 8$ and $y = -3$.

2. $4x^2 + 3x - 10 = 0$:

This equation doesn't seem to factor easily, so let's use the quadratic formula to solve for $x$:

The quadratic formula is given by:

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

In this equation, $a = 4$, $b = 3$, and $c = -10$. Substituting these values into the formula:

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

Simplifying the expression under the square root:

$x = \frac{-3 \pm \sqrt{9 + 160}}{8}$ $x = \frac{-3 \pm \sqrt{169}}{8}$ $x = \frac{-3 \pm 13}{8}$

So, we have two possible solutions for $x$:

$x = \frac{-3 + 13}{8} = \frac{10}{8} = \frac{5}{4}$

$x = \frac{-3 - 13}{8} = \frac{-16}{8} = -2$

The solutions for this equation are $x = \frac{5}{4}$ and $x = -2$.

In summary, the solutions for the first equation $y^2 - 5y - 24 = 0$ are $y = 8$ and $y = -3$, and the solutions for the second equation $4x^2 + 3x - 10 = 0$ are $x = \frac{5}{4}$ and $x = -2$.