42x=-9x^2-38

Solution for 42x=-9x^2-38 equation:

42x=-9x^2-38

We move all terms to the left:

42x-(-9x^2-38)=0

We get rid of parentheses

9x^2+42x+38=0

a = 9; b = 42; c = +38;
Δ = b2-4ac
Δ = 422-4·9·38
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{11}}{2*9}=\frac{-42-6\sqrt{11}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{11}}{2*9}=\frac{-42+6\sqrt{11}}{18}
$