18x^2+60x+42=0

Solution for 18x^2+60x+42=0 equation:

18x^2+60x+42=0

a = 18; b = 60; c = +42;
Δ = b2-4ac
Δ = 602-4·18·42
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-24}{2*18}=\frac{-84}{36}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+24}{2*18}=\frac{-36}{36}
=-1
$