x=12x^2-41x+35

Solution for x=12x^2-41x+35 equation:

x=12x^2-41x+35

We move all terms to the left:

x-(12x^2-41x+35)=0

We get rid of parentheses

-12x^2+x+41x-35=0

We add all the numbers together, and all the variables

-12x^2+42x-35=0

a = -12; b = 42; c = -35;
Δ = b2-4ac
Δ = 422-4·(-12)·(-35)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{21}}{2*-12}=\frac{-42-2\sqrt{21}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{21}}{2*-12}=\frac{-42+2\sqrt{21}}{-24}
$