9x^2+39x+42=0

Solution for 9x^2+39x+42=0 equation:

9x^2+39x+42=0

a = 9; b = 39; c = +42;
Δ = b2-4ac
Δ = 392-4·9·42
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-3}{2*9}=\frac{-42}{18}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+3}{2*9}=\frac{-36}{18}
=-2
$