-.05x^2+35=20

Solution for -.05x^2+35=20 equation:

-.05x^2+35=20

We move all terms to the left:

-.05x^2+35-(20)=0

We add all the numbers together, and all the variables

-0.05x^2+15=0

a = -0.05; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-0.05)·15
Δ = 3
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{3}}{2*-0.05}=\frac{0-\sqrt{3}}{-0.1}
=-\frac{\sqrt{}}{-0.1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{3}}{2*-0.05}=\frac{0+\sqrt{3}}{-0.1}
=\frac{\sqrt{}}{-0.1}
$