# 2y^2+21=13y please solve the equation in formula method

2
by saikrishnan

## Answers

2014-11-20T17:43:48+05:30
This method may be difficult but it was thought in ma school days
2y ^2 + 21 = 13y
2y^2 + 21 -13y
2y^2 - 13y + 21
multiply 2 with 21 we get 42
then the equation be y^2 - 13y + 42
let factorise we get
[( 2y - 7 ) ( 2y -6)]/ 2
[ (2y-7) 2( y-3) ]/2                          taking 2 common outside and divide with 2
(2y-7) ( y -3) = o
2y =7 ; y =3
y = 7/2 and y = 3
2014-11-20T18:11:05+05:30
3.5, 3, by using the quadratics equation: y= [-b +/- √(b²-4ac)]÷2aOkay,
if we were to substitute the values, then we would get:

the equation is 2y^2+21=13y right, so transpose to make a zero.
make it 2y^2-13y+21=0
this is in the form of ay^2+by+c=0, right?,
now, a=2, b=-13, c=21.

now, using the equation itself.
the equation is, y= [-b +/- √(b²-4ac)]÷2a.
so, y={-(-13)+/- √(-13²-4(2*21)}/4
this equals to {13+/-√(169-168) }/4
this is [13+/- 1] /4
so, this can either be [13+1]/4 or [13-1]/4
so the answers are 12/4 and 14/4, which are 3 and 3.5 respectively.
Okay, if wee were to substitute the values, then we would get:
the equation is 2y^2+21=13y right, so transpose to make a zero. make it 2y^2-13y+21=0 this is in the form of ay^2+by+c=0, right?, now, a=2, b=-13, c=21.
now, using the equation itself. the equation is, y= [-b +/- √(b²-4ac)]÷2a. so, y={-(-13)+/- √(-13²-4(2*21)}/4 this equals to {13+/-√(169-168) }/4 this is [13+/- 1] /4 so, this can either be [13+1]/4 or [13-1]/4 so the answers are 12/4 and 14/4, which are 3 and 3.5 respectively.
Dude, i wrote it again in the answer box.