12x^2-140x+300=0

Solution for 12x^2-140x+300=0 equation:

12x^2-140x+300=0

a = 12; b = -140; c = +300;
Δ = b2-4ac
Δ = -1402-4·12·300
Δ = 5200
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:

$\sqrt{\Delta}=\sqrt{5200}=\sqrt{400*13}=\sqrt{400}*\sqrt{13}=20\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-20\sqrt{13}}{2*12}=\frac{140-20\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+20\sqrt{13}}{2*12}=\frac{140+20\sqrt{13}}{24}
$