-2r(r+3)=-35

Solution for -2r(r+3)=-35 equation:

-2r(r+3)=-35

We move all terms to the left:

-2r(r+3)-(-35)=0

We add all the numbers together, and all the variables

-2r(r+3)+35=0

We multiply parentheses

-2r^2-6r+35=0

a = -2; b = -6; c = +35;
Δ = b2-4ac
Δ = -62-4·(-2)·35
Δ = 316
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{79}}{2*-2}=\frac{6-2\sqrt{79}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{79}}{2*-2}=\frac{6+2\sqrt{79}}{-4}
$