Y=-4.9x^2+20x-5

Solution for Y=-4.9x^2+20x-5 equation:

=-4.9Y^2+20Y-5

We move all terms to the left:

-(-4.9Y^2+20Y-5)=0

We get rid of parentheses

4.9Y^2-20Y+5=0

a = 4.9; b = -20; c = +5;
Δ = b2-4ac
Δ = -202-4·4.9·5
Δ = 302
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{-(-20)-\sqrt{302}}{2*4.9}=\frac{20-\sqrt{302}}{9.8}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{302}}{2*4.9}=\frac{20+\sqrt{302}}{9.8}
$