12r^2-5r-25=0

Solution for 12r^2-5r-25=0 equation:

12r^2-5r-25=0

a = 12; b = -5; c = -25;
Δ = b2-4ac
Δ = -52-4·12·(-25)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-35}{2*12}=\frac{-30}{24}
=-1+1/4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+35}{2*12}=\frac{40}{24}
=1+2/3
$