11r^2-15r+4=0

Solution for 11r^2-15r+4=0 equation:

11r^2-15r+4=0

a = 11; b = -15; c = +4;
Δ = b2-4ac
Δ = -152-4·11·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-7}{2*11}=\frac{8}{22}
=4/11
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+7}{2*11}=\frac{22}{22}
=1
$