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

1 Answer
Mar 31, 2017

Answer:

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

Explanation:

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.