10r^2+20r-2=18

Solution for 10r^2+20r-2=18 equation:

10r^2+20r-2=18

We move all terms to the left:

10r^2+20r-2-(18)=0

We add all the numbers together, and all the variables

10r^2+20r-20=0

a = 10; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·10·(-20)
Δ = 1200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{3}}{2*10}=\frac{-20-20\sqrt{3}}{20}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{3}}{2*10}=\frac{-20+20\sqrt{3}}{20}
$