How do you solve $2t^2 – 6t + 1 = 0$ using the quadratic formula?

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Answer 1 · 2015-10-14T06:41:40

The solutions are:
$color(blue)(t=((3+sqrt(7)))/2$

$color(blue)(t=((3-sqrt(7)))/2$

$2t^2-6t+1$

The equation is of the form $color(blue)(at^2+bt+c=0$ where:
$a=2, b=-6, c=1$

The Discriminant is given by:

$Delta=b^2-4*a*c$
$= (-6)^2-(4*2*1)$
$= 36- 8=28$

The solutions are found using the formula
$t=(-b+-sqrtDelta)/(2*a)$

$t = (-(-6)+-sqrt(28))/(2*2) = (6+-2sqrt(7))/4$

$(6+-2sqrt(7))/4 = (2(3+-sqrt(7)))/4$

$=(cancel2(3+-sqrt(7)))/cancel4$

$=((3+-sqrt(7)))/2$

The solutions are
$color(blue)(t=((3+sqrt(7)))/2$

$color(blue)(t=((3-sqrt(7)))/2$

How do you solve 2t^2 – 6t + 1 = 02t^2 – 6t + 1 = 0 using the quadratic formula?