3a^2+42a=-7

Solution for 3a^2+42a=-7 equation:

3a^2+42a=-7

We move all terms to the left:

3a^2+42a-(-7)=0

We add all the numbers together, and all the variables

3a^2+42a+7=0

a = 3; b = 42; c = +7;
Δ = b2-4ac
Δ = 422-4·3·7
Δ = 1680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-4\sqrt{105}}{2*3}=\frac{-42-4\sqrt{105}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+4\sqrt{105}}{2*3}=\frac{-42+4\sqrt{105}}{6}
$