12x^2-42x+36=0

Solution for 12x^2-42x+36=0 equation:

12x^2-42x+36=0

a = 12; b = -42; c = +36;
Δ = b2-4ac
Δ = -422-4·12·36
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-6}{2*12}=\frac{36}{24}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+6}{2*12}=\frac{48}{24}
=2
$