2r^2-23r+56=0

Solution for 2r^2-23r+56=0 equation:

2r^2-23r+56=0

a = 2; b = -23; c = +56;
Δ = b2-4ac
Δ = -232-4·2·56
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-9}{2*2}=\frac{14}{4}
=3+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+9}{2*2}=\frac{32}{4}
=8
$