42x^2+68x=-18

Solution for 42x^2+68x=-18 equation:

42x^2+68x=-18

We move all terms to the left:

42x^2+68x-(-18)=0

We add all the numbers together, and all the variables

42x^2+68x+18=0

a = 42; b = 68; c = +18;
Δ = b2-4ac
Δ = 682-4·42·18
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-40}{2*42}=\frac{-108}{84}
=-1+2/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+40}{2*42}=\frac{-28}{84}
=-1/3
$