3y^2-28y+65=0

Solution for 3y^2-28y+65=0 equation:

3y^2-28y+65=0

a = 3; b = -28; c = +65;
Δ = b2-4ac
Δ = -282-4·3·65
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-2}{2*3}=\frac{26}{6}
=4+1/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+2}{2*3}=\frac{30}{6}
=5
$