3y^2-3y-429/4=0

Solution for 3y^2-3y-429/4=0 equation:

3y^2-3y-429/4=0

We multiply all the terms by the denominator


3y^2*4-3y*4-429=0

Wy multiply elements

12y^2-12y-429=0

a = 12; b = -12; c = -429;
Δ = b2-4ac
Δ = -122-4·12·(-429)
Δ = 20736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20736}=144$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-144}{2*12}=\frac{-132}{24}
=-5+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+144}{2*12}=\frac{156}{24}
=6+1/2
$