10=-2.5+0.5r^2

Solution for 10=-2.5+0.5r^2 equation:

10=-2.5+0.5r^2

We move all terms to the left:

10-(-2.5+0.5r^2)=0

We get rid of parentheses

-0.5r^2+2.5+10=0

We add all the numbers together, and all the variables

-0.5r^2+12.5=0

a = -0.5; b = 0; c = +12.5;
Δ = b2-4ac
Δ = 02-4·(-0.5)·12.5
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-5}{2*-0.5}=\frac{-5}{-1}
=+5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+5}{2*-0.5}=\frac{5}{-1}
=-5
$