How do you find the roots, real and imaginary, of #y= -2x^2 + 15x + 22 # using the quadratic formula?

1 Answer
Jan 18, 2016

Answer:

at #y=0# #color(white)(.........)x= + 8.756 " or " -1.256#

Explanation:

Given: #color(white)(...)color(brown)(y=-2x^2+15x+22)#

Using standard for of # y=ax^2+bx+c=0#

Where: #color(white)(....)x=(-b+-sqrt(b^2-4ac))/(2a)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#a=-2#
#b=15#
#c=22#

Thus: #color(white)(.....)x=(-15+-sqrt(15^2-4(-2)(22)))/(2(-2)) #

#x= (-15+-sqrt(225 + 176))/(-4)#

#x=(-15+-sqrt(401))/(-4)#

But 401 is a prime number so unable to break it down further

Using #sqrt(401)=20.025# to 3 decimal places

#x= + 8.756 " or " -1.256#
Tony B