42x^2+67x+10=0

Solution for 42x^2+67x+10=0 equation:

42x^2+67x+10=0

a = 42; b = 67; c = +10;
Δ = b2-4ac
Δ = 672-4·42·10
Δ = 2809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2809}=53$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(67)-53}{2*42}=\frac{-120}{84}
=-1+3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(67)+53}{2*42}=\frac{-14}{84}
=-1/6
$