7r^2+36r+35=3

Solution for 7r^2+36r+35=3 equation:

7r^2+36r+35=3

We move all terms to the left:

7r^2+36r+35-(3)=0

We add all the numbers together, and all the variables

7r^2+36r+32=0

a = 7; b = 36; c = +32;
Δ = b2-4ac
Δ = 362-4·7·32
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{-(36)-20}{2*7}=\frac{-56}{14}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+20}{2*7}=\frac{-16}{14}
=-1+1/7
$