How do you solve $-2y^2+6y=-2$ ?

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Answer 1 · 2015-07-06T08:32:53

$color(blue)(y=(3+sqrt13)/(2) , y=(3-sqrt13)/(2$

$2y^2 -6y-2 =0$

The equation is of the form $color(blue)(ay^2+by+c=0$ where:

$color(blue)(a=2, b=-6, c=-2$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (-6)^2-(4*(2)*-2)$

$= 36 + 16=52$

The solution is found using the formula:

$y=(-b+-sqrtDelta)/(2*a)$

$y= (-(-6)+-sqrt(52))/(2*2) = (6+-sqrt(52))/4= (6+-2sqrt13)/4$

$y=(6+2sqrt13)/(4) , x=(6-2sqrt13)/(4$

$color(blue)(y=(3+sqrt13)/(2) , y=(3-sqrt13)/(2$

How do you solve -2y^2+6y=-2-2y^2+6y=-2?