r2+5r-126=0

Solution for r2+5r-126=0 equation:

r2+5r-126=0

We add all the numbers together, and all the variables

r^2+5r-126=0

a = 1; b = 5; c = -126;
Δ = b2-4ac
Δ = 52-4·1·(-126)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-23}{2*1}=\frac{-28}{2}
=-14
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+23}{2*1}=\frac{18}{2}
=9
$