How do you solve #p^2+7p+8=0# using the quadratic formula?

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Answer 1 · 2017-07-19T03:25:44

See a solution process below:

For #ax^2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

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

Substituting #1# for #a#; #7# for #b# and #8# for #c# gives:

#x = (-7 +- sqrt(7^2 - (4 * 1 * 8)))/(2 * 1)#

#x = (-7 +- sqrt(49 - 32))/2#

#x = (-7 +- sqrt(17))/2#

Or

#x = (-7 + sqrt(17))/2# and #x = (-7 - sqrt(17))/2#