(r+2)+(r2-13)=35

Solution for (r+2)+(r2-13)=35 equation:

(r+2)+(r2-13)=35

We move all terms to the left:

(r+2)+(r2-13)-(35)=0

We add all the numbers together, and all the variables

(+r^2-13)+(r+2)-35=0

We get rid of parentheses

r^2+r-13+2-35=0

We add all the numbers together, and all the variables

r^2+r-46=0

a = 1; b = 1; c = -46;
Δ = b2-4ac
Δ = 12-4·1·(-46)
Δ = 185
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{185}}{2*1}=\frac{-1-\sqrt{185}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{185}}{2*1}=\frac{-1+\sqrt{185}}{2}
$