0=12x^2-41x+35

Solution for 0=12x^2-41x+35 equation:

0=12x^2-41x+35

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -12; b = 41; c = -35;
Δ = b2-4ac
Δ = 412-4·(-12)·(-35)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-1}{2*-12}=\frac{-42}{-24}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+1}{2*-12}=\frac{-40}{-24}
=1+2/3
$