-3x^2-42x=10

Solution for -3x^2-42x=10 equation:

-3x^2-42x=10

We move all terms to the left:

-3x^2-42x-(10)=0

a = -3; b = -42; c = -10;
Δ = b2-4ac
Δ = -422-4·(-3)·(-10)
Δ = 1644
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:

$\sqrt{\Delta}=\sqrt{1644}=\sqrt{4*411}=\sqrt{4}*\sqrt{411}=2\sqrt{411}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{411}}{2*-3}=\frac{42-2\sqrt{411}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{411}}{2*-3}=\frac{42+2\sqrt{411}}{-6}
$