3x^2+x^2-38x+24=0

Solution for 3x^2+x^2-38x+24=0 equation:

3x^2+x^2-38x+24=0

We add all the numbers together, and all the variables

4x^2-38x+24=0

a = 4; b = -38; c = +24;
Δ = b2-4ac
Δ = -382-4·4·24
Δ = 1060
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:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{265}}{2*4}=\frac{38-2\sqrt{265}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{265}}{2*4}=\frac{38+2\sqrt{265}}{8}
$