Question

2y square+27y+13=0.Solve the quadratic equation by factorisation.

Answer 1

2y^2+26y+y+13=0
2y(y+13)+1(y+13)=0
(2y+1)(y+13)=0

=>2y+1=0
2y= -1
y= -1/2

=>y+13=0
y= -13

=>y=(-1/2) or (-13)

Hope it helps u....

Answer 2

$Q. \: 2 {y}^{2} + 27y + 13 = 0. Solve \: the \\ quadratic \: equation \: by \: factorisation.$

$SOLUTION:-$

$= > 2 {y}^{2} + 27y + 13 = 0 \\ \\ Splitting \: the \: middle \: term. \\ So, \: that \: sum \: of \: middle \: term \\ is \: equal \: to \: product \: of \: other \: two \\ terms \: i.e. \: 1st \: and \: 3rd \: term \\ \\ = > 2 {y}^{2} + 26y + y + 13 = 0\\ \\ Select \: the common \: term \\ \\ = > 2y(y + 13) + 1(y + 13) = 0\\ \\ = > (y + 13)(2y + 1) = 0 \\ \\ Put \: each \: term \: with \: zero \: to \: get \: value \: of \: y \\ \\ = > y + 13 = 0 \: \: \: \: \: \: \: \: \: 2y + 1 = 0 \\ \\ = > y = - 13\: \: \: \: \: \: \: \: \:y = - \frac{1}{2} \\ \\ = > y = - 13 \: or \: - \frac{1}{2}$

$Ans:$
$\boxed{y = - 13 \: or \: - \frac{1}{2} }$