3x2+42x=96

Solution for 3x2+42x=96 equation:

3x^2+42x=96

We move all terms to the left:

3x^2+42x-(96)=0

a = 3; b = 42; c = -96;
Δ = b2-4ac
Δ = 422-4·3·(-96)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-54}{2*3}=\frac{-96}{6}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+54}{2*3}=\frac{12}{6}
=2
$