20r^2+40r+18=3

Solution for 20r^2+40r+18=3 equation:

20r^2+40r+18=3

We move all terms to the left:

20r^2+40r+18-(3)=0

We add all the numbers together, and all the variables

20r^2+40r+15=0

a = 20; b = 40; c = +15;
Δ = b2-4ac
Δ = 402-4·20·15
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*20}=\frac{-60}{40}
=-1+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*20}=\frac{-20}{40}
=-1/2
$