r^2-34r+120=0

Solution for r^2-34r+120=0 equation:

r^2-34r+120=0

a = 1; b = -34; c = +120;
Δ = b2-4ac
Δ = -342-4·1·120
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-26}{2*1}=\frac{8}{2}
=4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+26}{2*1}=\frac{60}{2}
=30
$