28x^2-48x+9=0

Solution for 28x^2-48x+9=0 equation:

28x^2-48x+9=0

a = 28; b = -48; c = +9;
Δ = b2-4ac
Δ = -482-4·28·9
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-36}{2*28}=\frac{12}{56}
=3/14
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+36}{2*28}=\frac{84}{56}
=1+1/2
$