3x^2+48x=1260

Solution for 3x^2+48x=1260 equation:

3x^2+48x=1260

We move all terms to the left:

3x^2+48x-(1260)=0

a = 3; b = 48; c = -1260;
Δ = b2-4ac
Δ = 482-4·3·(-1260)
Δ = 17424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{17424}=132$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-132}{2*3}=\frac{-180}{6}
=-30
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+132}{2*3}=\frac{84}{6}
=14
$