x=-0.01x^2+x+62

Solution for x=-0.01x^2+x+62 equation:

x=-0.01x^2+x+62

We move all terms to the left:

x-(-0.01x^2+x+62)=0

We get rid of parentheses

0.01x^2-x+x-62=0

We add all the numbers together, and all the variables

0.01x^2-62=0

a = 0.01; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·0.01·(-62)
Δ = 2.48
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{2.48}}{2*0.01}=\frac{0-\sqrt{2.48}}{0.02}
=-\frac{\sqrt{}}{0.02}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{2.48}}{2*0.01}=\frac{0+\sqrt{2.48}}{0.02}
=\frac{\sqrt{}}{0.02}
$