6(x-5)(x+2)=48

Solution for 6(x-5)(x+2)=48 equation:

6(x-5)(x+2)=48

We move all terms to the left:

6(x-5)(x+2)-(48)=0

We multiply parentheses ..

6(+x^2+2x-5x-10)-48=0

We multiply parentheses

6x^2+12x-30x-60-48=0

We add all the numbers together, and all the variables

6x^2-18x-108=0

a = 6; b = -18; c = -108;
Δ = b2-4ac
Δ = -182-4·6·(-108)
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{-(-18)-54}{2*6}=\frac{-36}{12}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+54}{2*6}=\frac{72}{12}
=6
$