Y=-3x^2+10

Solution for Y=-3x^2+10 equation:

=-3Y^2+10

We move all terms to the left:

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

We get rid of parentheses

3Y^2-10=0

a = 3; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·3·(-10)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{30}}{2*3}=\frac{0-2\sqrt{30}}{6}
=-\frac{2\sqrt{30}}{6}
=-\frac{\sqrt{30}}{3}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{30}}{2*3}=\frac{0+2\sqrt{30}}{6}
=\frac{2\sqrt{30}}{6}
=\frac{\sqrt{30}}{3}
$