Question

5x^2-22x-15=0 плиззз​

Answer 1

$\displaystyle \tt 5x^2-22x-15=0\\\\a=5,\:\:\:b=-22,\:\:\:c=-15\\\\\displaystyle \tt D=b^2-4ac=(-22)^2-4\cdot5\cdot(-15)=484+300=784\\\\ \displaystyle \tt \sqrt{D}=\sqrt{784}=28\\\\\displaystyle \tt \bold{x_1}}=\frac{-b+\sqrt{D}}{2a}=\frac{22+28}{2\cdot5}=\frac{50}{10}=5\\\\ \displaystyle \tt \bold{x_2}}=\frac{-b+\sqrt{D}}{2a}=\frac{22-28}{2\cdot5}=\frac{-6}{10}=-0,6$

Answer 2

To solve the quadratic equation 5x^2 - 22x - 15 = 0, you can use the quadratic formula:

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

For the equation 5x^2 - 22x - 15 = 0:

a = 5, b = -22, and c = -15.

Now, plug these values into the quadratic formula:

x = (22 ± √((-22)^2 - 4 * 5 * (-15))) / 2 * 5 x = (22 ± √(484 + 300)) / 10 x = (22 ± √784) / 10 x = (22 ± 28) / 10

Now, consider both the plus and minus options:

1. When x = (22 + 28) / 10: x = 50 / 10 x = 5

2. When x = (22 - 28) / 10: x = -6 / 10 x = -0.6

So, the solutions to the equation 5x^2 - 22x - 15 = 0 are x = 5 and x = -0.6.