y=-3y2+8y-4=

Solution for y=-3y2+8y-4= equation:

y=-3y^2+8y-4=

We move all terms to the left:

y-(-3y^2+8y-4)=0

We get rid of parentheses

3y^2-8y+y+4=0

We add all the numbers together, and all the variables

3y^2-7y+4=0

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

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*3}=\frac{6}{6}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*3}=\frac{8}{6}
=1+1/3
$