Y=-3x^2+42

Solution for Y=-3x^2+42 equation:

=-3Y^2+42

We move all terms to the left:

-(-3Y^2+42)=0

We get rid of parentheses

3Y^2-42=0

a = 3; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·3·(-42)
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*3}=\frac{0-6\sqrt{14}}{6}
=-\frac{6\sqrt{14}}{6}
=-\sqrt{14}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*3}=\frac{0+6\sqrt{14}}{6}
=\frac{6\sqrt{14}}{6}
=\sqrt{14}
$