21x^2+41x+18=0

Solution for 21x^2+41x+18=0 equation:

21x^2+41x+18=0

a = 21; b = 41; c = +18;
Δ = b2-4ac
Δ = 412-4·21·18
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-13}{2*21}=\frac{-54}{42}
=-1+2/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+13}{2*21}=\frac{-28}{42}
=-2/3
$