15r^2+56r+20=-5r^2+15r

Solution for 15r^2+56r+20=-5r^2+15r equation:

15r^2+56r+20=-5r^2+15r

We move all terms to the left:

15r^2+56r+20-(-5r^2+15r)=0

We get rid of parentheses

15r^2+5r^2-15r+56r+20=0

We add all the numbers together, and all the variables

20r^2+41r+20=0

a = 20; b = 41; c = +20;
Δ = b2-4ac
Δ = 412-4·20·20
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-9}{2*20}=\frac{-50}{40}
=-1+1/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+9}{2*20}=\frac{-32}{40}
=-4/5
$