Posted by Lynda on Tuesday, July 26, 2011 at 12:42pm.
______
/b24ac; b=3, a=1, c=2
What would the solution be?

algebra  Anonymous, Tuesday, July 26, 2011 at 12:54pm
b^24ac=98=1
X1=(bsqrt(1))/2a=(31)/(2)=2
X2=(b+sqrt(1))/2a=(3+1)/(2)=1