How do you solve #x^2-10x+18=0# using the quadratic formula?

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Answer 1 · 2017-03-31T00:13:45

You simply put the coefficients of the equation into the formula and solve it algebraically.

For #ax^2 + bx + c = 0#, the values of x which are the solutions of the equation are given by: #x = (−b±√[b^2−4ac])/(2a)#

In this case, a = 1, b = -10 and c = 18

#x = (−(-10) ± sqrt[(-10^2 − 4*1*18]))/(2*1)#

#x = (10 ± sqrt[(100 – 72)])/2# ; #x = (10 ± sqrt[28])/2#

#x = (10 ± 5.29)/2#

x = 2.35, and x = 7.645

CHECK: 7.645^2 – 10*7.645 + 18 = 0 ; 58.446 – 76.45 + 18 = 0 ; 0 = 0 correct.
See http://www.purplemath.com/modules/quadform.htm for instructions.