x^2=7x-9
2007-12-20 10:36:10 · 5 answers · asked by shy one 2 in Science & Mathematics ➔ Mathematics
x^2 - 7x + 9 = 0
7 +/- sqrt(49 - 36)/ 2 = 7 +/- sqrt(13)/2
2007-12-20 10:39:47 · answer #1 · answered by Blueberry Man 5 · 0⤊ 0⤋
make them on 1 side
0 = -x^2 +7x - 9
-7 + or - sqrt 7^2 - 4(-1)(-9) over -2
2007-12-20 11:22:47 · answer #2 · answered by ViewtifulJoe 4 · 0⤊ 0⤋
x² = 7x - 9
x² - 7x + 9 = 0
x = 7/2 ± [â(49 - 36)]/2
x = 7/2 ± [â13]/2
2007-12-20 10:40:07 · answer #3 · answered by Philo 7 · 0⤊ 0⤋
x^2-7x+9=0
This equation is of form ax2+bx+c
a = 1 b = -7 c = 9
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[7 +/-sqrt(-7^2-4(1)(9)]/(2)(1)
discriminant is b^2-4ac =13
x=[7 +sqrt(13)] / (2)(1)
x=[7 -sqrt(13)] / (2)(1)
x=[7+3.605] / 2
x=[7-3.605] / 2
The roots are 5.303 and 1.697
2007-12-20 10:39:56 · answer #4 · answered by cidyah 7 · 0⤊ 0⤋
-x^2+7x-9
-x62=a
7x=b
-9=c
put # in and solve:
x1=-b+[squ. root of (b^2)-4ac] / 2a
x2=-b-[squ. root of (b^2)-4ac] / 2a
there will be 2 answer x1 and x2
2007-12-20 10:41:15 · answer #5 · answered by jenn. 2 · 0⤊ 0⤋