6a^2+41a+63=0

Solution for 6a^2+41a+63=0 equation:

6a^2+41a+63=0

a = 6; b = 41; c = +63;
Δ = b2-4ac
Δ = 412-4·6·63
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-13}{2*6}=\frac{-54}{12}
=-4+1/2
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+13}{2*6}=\frac{-28}{12}
=-2+1/3
$