12x2+41x+35=0

Solution for 12x2+41x+35=0 equation:

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
$