r2-15r+44=0

Solution for r2-15r+44=0 equation:

r2-15r+44=0

We add all the numbers together, and all the variables

r^2-15r+44=0

a = 1; b = -15; c = +44;
Δ = b2-4ac
Δ = -152-4·1·44
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-7}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+7}{2*1}=\frac{22}{2}
=11
$