23.75r^2-68r+48=0

Solution for 23.75r^2-68r+48=0 equation:

23.75r^2-68r+48=0

a = 23.75; b = -68; c = +48;
Δ = b2-4ac
Δ = -682-4·23.75·48
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-8}{2*23.75}=\frac{60}{47.5}
=1+8.3333333333333/31.666666666667
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+8}{2*23.75}=\frac{76}{47.5}
=1+19/31.666666666667
$