How do you solve #x^2-2x+5=0# using the quadratic formula?

Question

Jace Anderson · Accepted Answer

undefined To solve the quadratic equation $x^2 - 2x + 5 = 0$ using the quadratic formula, first identify the coefficients $a$, $b$, and $c$, where $a = 1$, $b = -2$, and $c = 5$. Then, substitute these values into the quadratic formula $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$. After substitution, calculate the discriminant $b^2 - 4ac$. If the discriminant is negative, the equation has no real solutions. If it's zero, there is one real solution. If it's positive, there are two real solutions. Finally, use the formula to find the values of $x$ by substituting the discriminant and coefficients into the formula and solving for $x$.

Avery Alexander · Answer

The solutions are:
# x= (2+4i)/2#

# x= (2-4i)/2# #x^2 - 2x+5=0# The equation is of the form #color(blue)(ax^2+bx+c=0# where: #a=1, b=-2, c=5# The Discriminant is given by: #Delta=b^2-4*a*c# # = (-2)^2-(4*1*5)# # = 4 -20# #=-16# The solutions are found using the formula: #x=(-b+-sqrtDelta)/(2*a)# #x = (-(-2)+-sqrt(-16))/(2*1) = (2+-sqrt(-16))/2# # = (2+-sqrt(-1 xx 16))/2# # = (2+-4(i))/2# The solutions are: # x= (2+4i)/2# # x= (2-4i)/2#