How do you factor the expression $8y²-28y-60$ ?

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Answer 1 · 2018-06-30T15:49:05

$4(y-5)(3+2y)$

At first we Can factor $4(2y^2-7y-15)$ since
$8=2*4$
$28=7*4$
and
$60=15*4$

so we get

$4(2y^2-7y-15)$
now we solve the equation

$2y^2-7y-15=0$
dividing by $2$
$y^2-7/2y-15/2=0$
by the quadratic formula we get

$y_(1,2)=7/4pmsqrt(49/16+120/16)$
$y_(1,2)=7/4pmsqrt(169/16)$
so we get

$y_(1,2)=7/4pm13/4$
this gives

$y_1=5$
$y_2=6/4=3/2$

so we get

$4(y-5)(2y+3)$

How do you factor the expression 8y²-28y-608y²-28y-60?