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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+2r^2-13r+4r-26)-35=0

We get rid of parentheses

2r^2-13r+4r-26-35=0

We add all the numbers together, and all the variables

2r^2-9r-61=0

a = 2; b = -9; c = -61;
Δ = b2-4ac
Δ = -92-4·2·(-61)
Δ = 569
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{-(-9)-\sqrt{569}}{2*2}=\frac{9-\sqrt{569}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{569}}{2*2}=\frac{9+\sqrt{569}}{4}
$