How do you use the quadratic formula to find both solutions to the quadratic equation #x^2-3x=-10#?

Question

Savannah Anderson · Accepted Answer

Solve #y = x^2 - 3x + 10 = 0# #D = d^2 = b^2 - 4ac = 9 - 40 = -31 < 0# --> #d = +- isqrt31# There are no real roots. There are 2 complex roots. #x = 3/2 +-( isqrt31)/2#

Avery Alexander · Answer

#x^2-3x=-10# has only complex roots #x = (3+-i sqrt(31))/2#

#x^2-3x=10# has roots #x=5# and #x=-2# I'm not sure the sign of #10# on the right hand side is correct, so let's deal with both cases: Case 1 #x^2-3x=-10# Add #10# to both sides to get #x^2-3x+10 = 0# Let #f(x) = x^2-3x+10#. This is of the form #ax^2+bx+c# with #a=1#, #b=-3# and #c=10#. The discriminant #Delta# is given by the formula: #Delta = b^2 - 4ac = (-3)^2-(4xx1xx10) = 9-40 = -31# Since this is negative #f(x) = 0# has two distinct complex roots, given by the formula: #x = (-b +-sqrt(Delta))/(2a) = (3+-i sqrt(31))/2# Case 2 #x^2-3x=10# Subtract #10# from both sides to get #x^2-3x-10 = 0# Let #f(x) = x^2-3x-10# This is of the form #ax^2+bx+c# with #a=1#, #b=-3# and #c=-10# The discriminant #Delta# is given by the formula: #Delta = b^2 - 4ac = (-3)^2-(4xx1xx-10) = 9+40 = 49 = 7^2# This is positive and a perfect square, so the roots of #f(x) = 0# are distinct rational real numbers, given by the formula: #x = (-b+-sqrt(Delta))/(2a) = (3+-7)/2# That is #x = -4/2 = -2# and #x = 10/2 = 5#

Hazel Anglin · Answer

undefined To use the quadratic formula, which is $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $, for the quadratic equation $ x^2 - 3x = -10 $, you need to identify $ a $, $ b $, and $ c $ from the equation $ ax^2 + bx + c = 0 $. In this case, $ a = 1 $, $ b = -3 $, and $ c = -10 $. Substitute these values into the quadratic formula and solve for $ x $. You'll get two solutions.