0=3x^2+25x+42

Solution for 0=3x^2+25x+42 equation:

0=3x^2+25x+42

We move all terms to the left:

0-(3x^2+25x+42)=0

We add all the numbers together, and all the variables

-(3x^2+25x+42)=0

We get rid of parentheses

-3x^2-25x-42=0

a = -3; b = -25; c = -42;
Δ = b2-4ac
Δ = -252-4·(-3)·(-42)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-11}{2*-3}=\frac{14}{-6}
=-2+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+11}{2*-3}=\frac{36}{-6}
=-6
$