How do you solve $2y^2 – 6y – 8 = 0$ using the quadratic formula?

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Answer 1 · 2016-09-17T16:49:34

$y=-1 "; " y=4$

Just use $y$ instead of $x$

$color(brown)("Given: "x=2y^2-6y-8=0)$

$a=2"; "b=-6"; "c=-8$

$x=ay^2+by+c$

Where: $" "y=(-b+-sqrt(b^2-4ac))/(2a)$

$=> y=(+6+-sqrt(6^2-4(2)(-8)))/(2(2))$

$y= 6/4+- sqrt(100)/4$

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$color(brown)(y=6/4+10/4=16/4=4)$

$color(brown)(y=6/4-10/4 = -4/4=-1)$

How do you solve 2y^2 – 6y – 8 = 02y^2 – 6y – 8 = 0 using the quadratic formula?