Question

4x^2-19x+12=0 help pls

Answer 1

A=4, b = -19, c=12
D = (-19)^2 - 4*4*12 = 169
x1 = (-(-19) + sqrt(169))/2*4 = (19+13)/8 = 4
x2 = (-(-19) - sqrt(169))/2*4 = (19-13)/8 = 0.75

Ответ: 4 и 0ю75

Answer 2

4x^2-19x+12=0
a=4
b=-19
c=12
D=b^2-4ac=(-19)^2-4*4*12=361-192=169
D>0
Квадратное уравнение имеет два квадратных корня.
х(1)=-b+13/2*a=19+13/2*4=32/8=4
x(2)=-b-13/2*a=19-13/2*4=6/8=3/4=0.75
Ответ: х(1)=4, х(2)=0.75

Answer 3

Solving the Quadratic Equation 4x^2 - 19x + 12 = 0

To solve the quadratic equation 4x^2 - 19x + 12 = 0, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's calculate the solutions for the given equation using this formula.

Applying the Quadratic Formula

For the equation 4x^2 - 19x + 12 = 0: - a = 4 - b = -19 - c = 12

Using the quadratic formula: x = (-(-19) ± √((-19)^2 - 4*4*12)) / (2*4)

x = (19 ± √(361 - 192)) / 8

x = (19 ± √169) / 8

x = (19 ± 13) / 8

So, the solutions for the given equation are: x1 = (19 + 13) / 8 = 32 / 8 = 4 x2 = (19 - 13) / 8 = 6 / 8 = 3/4

Therefore, the solutions for the equation 4x^2 - 19x + 12 = 0 are: x = 4 and x = 3/4.

This is the result obtained by applying the quadratic formula to the given equation.