Y=-5t^2+15+50

Solution for Y=-5t^2+15+50 equation:

=-5Y^2+15+50

We move all terms to the left:

-(-5Y^2+15+50)=0

We get rid of parentheses

5Y^2-15-50=0

We add all the numbers together, and all the variables

5Y^2-65=0

a = 5; b = 0; c = -65;
Δ = b2-4ac
Δ = 02-4·5·(-65)
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{13}}{2*5}=\frac{0-10\sqrt{13}}{10}
=-\frac{10\sqrt{13}}{10}
=-\sqrt{13}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{13}}{2*5}=\frac{0+10\sqrt{13}}{10}
=\frac{10\sqrt{13}}{10}
=\sqrt{13}
$