How do you solve #y = x^2 - 16x + 58# using the quadratic formula?

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Answer 1 · 2018-03-06T03:32:37

#x=8+-sqrt(6)#

For every #Ax^2+Bx+C=0#,

#x=[-B+-sqrt(B^2-4AC)]/(2A)#

For the equation #x^2-16x+58=0#

#x=[-(-16)+-sqrt((-16)^2-4(1)(58))]/(2(1))#

#x=[16+-sqrt(256-232)]/2#

#x=[16+-sqrt(24)]/2#

#x=[16+-2sqrt(6)]/2#

#x=8+-sqrt(6)#

Note: the sign #+-# means that both addition and subtraction gives answers, in this case, the two roots are #8+sqrt6# and #8-sqrt6#.

Answer 2 · 2018-03-06T03:45:06

#8+-sqrt(6)#

which would mean #8+sqrt(6)# and #8-sqrt(6)# are your two solutions.

Alright, so our quadratic formula is

#x_(1,2) = (-b +- sqrt(b^2-4ac))/(2a)#

So our #a#, #b#, and #c# values are all the leading coefficients, so #(1,-16,58)#

Next step: insert the numbers into the formula (double negative makes the #16# positive)

#x_(1,2) = (-(-16)+- sqrt(-16^2-4(1)(58)))/(2(1))#

#x_(1,2) = (8+-sqrt(24))/(2)#

Now we can simplify our square root by doing

#sqrt(24) = 2sqrt(6)#

The two's will now cancel out on the top and bottom and we will be left with

#x_(1,2) = 8+- sqrt(6)#