Y=-3x^2+x+12

Solution for Y=-3x^2+x+12 equation:

=-3Y^2+Y+12

We move all terms to the left:

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

We get rid of parentheses

3Y^2-Y-12=0

We add all the numbers together, and all the variables

3Y^2-1Y-12=0

a = 3; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·3·(-12)
Δ = 145
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{-(-1)-\sqrt{145}}{2*3}=\frac{1-\sqrt{145}}{6}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{145}}{2*3}=\frac{1+\sqrt{145}}{6}
$