y^=3,2y-10y^2

Solution for y^=3,2y-10y^2 equation:

y^=3.2y-10y^2

We move all terms to the left:

y^-(3.2y-10y^2)=0

We add all the numbers together, and all the variables

-(3.2y-10y^2)+y=0

We get rid of parentheses

10y^2-3.2y+y=0

We add all the numbers together, and all the variables

10y^2-2.2y=0

a = 10; b = -2.2; c = 0;
Δ = b2-4ac
Δ = -2.22-4·10·0
Δ = 4.84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2.2)-\sqrt{4.84}}{2*10}=\frac{2.2-\sqrt{4.84}}{20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.2)+\sqrt{4.84}}{2*10}=\frac{2.2+\sqrt{4.84}}{20}
$