3x^2-x^2-40x+48=0

Solution for 3x^2-x^2-40x+48=0 equation:

3x^2-x^2-40x+48=0

We add all the numbers together, and all the variables

2x^2-40x+48=0

a = 2; b = -40; c = +48;
Δ = b2-4ac
Δ = -402-4·2·48
Δ = 1216
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:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{19}}{2*2}=\frac{40-8\sqrt{19}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{19}}{2*2}=\frac{40+8\sqrt{19}}{4}
$