x^2=20
x^2+3x+12
2x^2+4x-3=0
2007-08-01 11:14:38 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2=20
x = +/- sqrt(20)
x = +/- 2 * sqrt(5)
x^2+3x+12
has no real solutions -- cannot be factored except using imaginary numbers -- since the discriminant (3^2 - 4*1*12 = -39) is not positive.
2x^2+4x-3=0
x = (-4 +/- sqrt(4^2 + 4*2*3))/(2*2)
x = -1 +/- sqrt(40)/4
x = -1 +/- sqrt(10)/2
2007-08-01 11:29:47 · answer #1 · answered by McFate 7 · 0⤊ 0⤋
x = +/- sqrt(20) = +/- 2sqrt(5).
Provided the expression equals 0:
x = ( - 3 +/- sqrt(3^2 - 48) ) / 2
= ( - 3 +/- i sqrt(39) ) / 2.
x = ( - 4 +/- sqrt(16 + 24) ) / 4
= ( - 2 +/- sqrt(10) ) / 2.
2007-08-01 18:31:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋
x^2=20
x = plus or minus the sqrt of 20
use quadratic formula to solve for x for the rest you know
x= -b +- sqrt(b squared - 4ac)/(2a)
2007-08-01 18:27:19 · answer #3 · answered by π∑∞∫questionqueen 3 · 0⤊ 0⤋