Question

5t²+20t−(t+4)=0 x1= x2=

Answer 1

$\displaystyle \tt 5t^{2}+20t-(t+4)=0$

$\displaystyle \tt 5t^{2}+20t-t-4=0$

$\displaystyle \tt 5t^2+19t-4=0$

$\displaystyle \tt D=b^{2}-4ac=19^2-4*5*(-4)=441$

$\displaystyle \tt t=\frac{-b-\sqrt{D} }{2a}=\frac{-19-21}{2*5}=-\frac{40}{10}=\bold\bold-\bold4$

$\displaystyle \tt t=\frac{-b+\sqrt{D} }{2a}=\frac{-19+21}{2*5}=\frac{2}{10}=\frac{\bold1}{\bold5} \bold$

Answer 2

To find the solutions for the quadratic equation 5t² + 20t - (t + 4) = 0, we can first simplify the equation:

5t² + 20t - (t + 4) = 0 5t² + 20t - t - 4 = 0 5t² + 19t - 4 = 0

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

The quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic equation ax² + bx + c = 0.

In our case, a = 5, b = 19, and c = -4. Now, let's plug these values into the formula to find x1 and x2:

x1 = (-19 + √(19² - 4 * 5 * -4)) / 2 * 5 x2 = (-19 - √(19² - 4 * 5 * -4)) / 2 * 5

Now, calculate the values:

x1 = (-19 + √(361 + 80)) / 10 x2 = (-19 - √(361 + 80)) / 10

x1 = (-19 + √441) / 10 x2 = (-19 - √441) / 10

x1 = (-19 + 21) / 10 x2 = (-19 - 21) / 10

x1 = 2 / 10 x2 = -40 / 10

x1 = 0.2 x2 = -4

So, the solutions to the quadratic equation are: x1 = 0.2 x2 = -4