Question

Решить дискриминант -t^2+5t-4

Answer 1

$-t^2+5t-4=0\\\\D=b^2-4ac=5^2-4\cdot (-1)\cdot (-4)=25-16=9\; ,\; \; \underline {D=9}\\\\x_{1,2}=\frac{-b\pm \sqrt{D}}{2a}= \frac{-5\pm \sqrt9}{2\cdot (-1)}=\frac{-5\pm 3}{-2}\\\\x_1=\frac{-5-3}{-2}=4\; ,\; \; x_2=\frac{-5+3}{-2}=1$

Answer 2

-t^2+5t-4=0; D=5^2-4*(-1)*(-4)=25-16=9; t1=(-5-3)/(-2), t2=(-5+3)/(-2). t1=4, t2=1.

Answer 3

To find the discriminant of a quadratic equation in the form of $at^2 + bt + c = 0$, where $a$, $b$, and $c$ are constants, you need to use the formula for the discriminant:

$D = b^2 - 4ac$

In your equation, $a = -1$, $b = 5$, and $c = -4$. Now, plug these values into the formula to find the discriminant:

$D = (5)^2 - 4(-1)(-4)$ $D = 25 - 16$ $D = 9$

So, the discriminant of the quadratic equation $t^2 - 5t - 4 = 0$ is $9$.