the roots of the equation: (x^2)-3x+7=0 are?

Quadratic Equation:
x² - 3x + 7 = 0
a = 1, b = -3, c = 7

x = [-b ± √(b² - 4ac)]/2a
x = [-(-3) ± √((-3)² - 4(1)(7))]/2(1)
x = [3 ± √(9 - 28)]/2
x = [3 ± √(-19)]/2
x = [3 ± √(-1*19)]/2
x = (3 ± √-1√19)/2
x = (3 ± i√19)/2

x = (3 + i√19)/2 = 1.5 + 2.1794i
x = (3 - i√19)/2 = 1.5 - 2.1794i


Completing the Square:
x² - 3x + 7 = 0
x² - 3x = -7
x² - 3x + (9/4) = -7 + (9/4)
(x - (3/2))(x - (3/2)) = (-28/4) + (9/4)
(x - (3/2))² = -19/4
x - (3/2) = ±√(-19/4)
x - (3/2) = ±(√-19)/√4
x - (3/2) = ±(√-1*19)/2
x - (3/2) = ±(i√19)/2
x = (3/2) ± (i√19)/2
x = (3 ± i√19)/2

x = (3 + i√19)/2 = 1.5 + 2.1794i
x = (3 - i√19)/2 = 1.5 - 2.1794i

x^2 - 3x + 7 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -3
c = 7

x = [3 ±√(9 - 28)]/2
x = [3 ±√(-19)]/2
x = [3 ±i√19]/2

∴ x = [3 ±i√19]/2

x = [ - 3 ± √ (9 - 28) ] / 2
x = [ - 3 ± √ (- 19) ] / 2
x = [ - 3 ± i √19 ] / 2

x² - 3x + 7 = 0
x = {3 ± √[3² - 4(1)(7)]}/2
x = {3 ± √-19}/2
x = (3 ± i√19)/2

x= (3±i√19)/2

Compliments of the quadratic formula.

1.5 - 2.179i
1.5 + 2.179i

using the quadratic equation, b squared minus 4ac must be greater than 0 for there to be real roots (you cant have a negative square root). b squared minus 4ac is equal to 9-28=-19, which means that there are no real roots.

You tell me and we will both know.

no real roots, Y never equals 0

1,7, 1/7