r2+6r-16=0

Solution for r2+6r-16=0 equation:

r2+6r-16=0

We add all the numbers together, and all the variables

r^2+6r-16=0

a = 1; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*1}=\frac{-16}{2}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*1}=\frac{4}{2}
=2
$