How do you solve #H= -0.5x^2 + x + 4# using the quadratic formula?

Question

Abigail Allen · Accepted Answer

undefined To solve the equation H = -0.5x^2 + x + 4 using the quadratic formula, where ax^2 + bx + c = 0:

1. Identify the coefficients:
   a = -0.5
   b = 1
   c = 4

2. Plug the coefficients into the quadratic formula:
   x = [-b ± √(b^2 - 4ac)] / (2a)

3. Substitute the coefficients into the formula:
   x = [-(1) ± √((1)^2 - 4(-0.5)(4))] / (2(-0.5))

4. Simplify the expression inside the square root:
   x = [-(1) ± √(1 + 8)] / (-1)

5. Further simplify the expression:
   x = [-(1) ± √(9)] / (-1)

6. Calculate the square root of 9:
   x = [-(1) ± 3] / (-1)

7. Determine the two possible solutions by considering both the addition and subtraction:
   x₁ = (1 + 3) / (-1) = 4 / (-1) = -4
   x₂ = (1 - 3) / (-1) = -2 / (-1) = 2

Therefore, the solutions to the equation are x = -4 and x = 2.

Nathan Axel · Answer

The answer #x_1=1+sqrt3# and #x_2=1-sqrt3#

show the steps below

#-0.5x^2+x+4=0#

Hit the two side by #-1/0.5#

#x^2-x/0.5-2=0#

#-0.5=-1/2#

#x^2-2x-2=0#

You can solve it by using the law:

#x_1=(-b+sqrt(b^2-4ac))/(2a)#

#x_2=(-b-sqrt(b^2-4ac))/(2a)#

in our formula the values of a b and c equal

#a=1 ,,,,,, b=-2 ,,,,, c=-2#

#x_1=(2+sqrt(4+8))/2#

#x_1=1+sqrt3#

in the same way #x_2# equal

#x_2=1-sqrt3#