Y=3x2-12x+10

Solution for Y=3x2-12x+10 equation:

=3Y^2-12Y+10

We move all terms to the left:

-(3Y^2-12Y+10)=0

We get rid of parentheses

-3Y^2+12Y-10=0

a = -3; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·(-3)·(-10)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{6}}{2*-3}=\frac{-12-2\sqrt{6}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{6}}{2*-3}=\frac{-12+2\sqrt{6}}{-6}
$