3y2+-7y+4=0

Solution for 3y2+-7y+4=0 equation:

3y^2+-7y+4=0

We add all the numbers together, and all the variables

3y^2-7y=0

a = 3; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·3·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*3}=\frac{14}{6}
=2+1/3
$