r^2+8r-34=50

Solution for r^2+8r-34=50 equation:

r^2+8r-34=50

We move all terms to the left:

r^2+8r-34-(50)=0

We add all the numbers together, and all the variables

r^2+8r-84=0

a = 1; b = 8; c = -84;
Δ = b2-4ac
Δ = 82-4·1·(-84)
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{-(8)-20}{2*1}=\frac{-28}{2}
=-14
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*1}=\frac{12}{2}
=6
$