3x^2+120=42x

Solution for 3x^2+120=42x equation:

3x^2+120=42x

We move all terms to the left:

3x^2+120-(42x)=0

a = 3; b = -42; c = +120;
Δ = b2-4ac
Δ = -422-4·3·120
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-18}{2*3}=\frac{24}{6}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+18}{2*3}=\frac{60}{6}
=10
$