10r^2-51r+27=0

Solution for 10r^2-51r+27=0 equation:

10r^2-51r+27=0

a = 10; b = -51; c = +27;
Δ = b2-4ac
Δ = -512-4·10·27
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-51)-39}{2*10}=\frac{12}{20}
=3/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-51)+39}{2*10}=\frac{90}{20}
=4+1/2
$