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 = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{66}}{2*-3}=\frac{12-2\sqrt{66}}{-6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{66}}{2*-3}=\frac{12+2\sqrt{66}}{-6}
$