r2-18r+56=0

Solution for r2-18r+56=0 equation:

r2-18r+56=0

We add all the numbers together, and all the variables

r^2-18r+56=0

a = 1; b = -18; c = +56;
Δ = b2-4ac
Δ = -182-4·1·56
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{-(-18)-10}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+10}{2*1}=\frac{28}{2}
=14
$