What are the zeros #-2x^2-15x+y+22=0#?

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Answer 1 · 2018-05-21T02:40:59

#x=(-15+sqrt401)/4#, #(-15-sqrt401)/4#

Given:

#-2x^2-15x+y+22=0#

Subtract #y# from both sides.

#-2x^2-15x+22=-y#

Multiply both sides by #-1#. This will reverse the signs.

#2x^2+15x-22=y#

Switch sides.

#y=2x^2+15x-22#

This is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=2#, #b=15#, #c=-22#

The roots are the x-intercepts, which are the values for #x# when #y=0#.

Substitute #0# for #y#.

#0=2x^2+15x-22#

Solve for #x# using the quadratic formula:

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

Plug the known values into the equation.

#x=(-15+-sqrt(15^2-4*2*-22))/(2*2)#

#x=(-15+-sqrt(401))/4# #larr# #401# is a prime number

Roots

#x=(-15+sqrt401)/4#, #(-15-sqrt401)/4#

Approximate roots

#x~~2.56,# #-8.756#

graph{y=2x^2+15x-22 [-11.09, 11.41, -8.775, 2.475]}