12x^2-252x+902=0

Solution for 12x^2-252x+902=0 equation:

12x^2-252x+902=0

a = 12; b = -252; c = +902;
Δ = b2-4ac
Δ = -2522-4·12·902
Δ = 20208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20208}=\sqrt{16*1263}=\sqrt{16}*\sqrt{1263}=4\sqrt{1263}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-252)-4\sqrt{1263}}{2*12}=\frac{252-4\sqrt{1263}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-252)+4\sqrt{1263}}{2*12}=\frac{252+4\sqrt{1263}}{24}
$