5x-20=3(x-4)24x+12

Solution for 5x-20=3(x-4)24x+12 equation:

5x-20=3(x-4)24x+12

We move all terms to the left:

5x-20-(3(x-4)24x+12)=0


We calculate terms in parentheses: -(3(x-4)24x+12), so:

3(x-4)24x+12

We multiply parentheses

72x^2-288x+12

Back to the equation:

-(72x^2-288x+12)


We get rid of parentheses

-72x^2+5x+288x-12-20=0

We add all the numbers together, and all the variables

-72x^2+293x-32=0

a = -72; b = 293; c = -32;
Δ = b2-4ac
Δ = 2932-4·(-72)·(-32)
Δ = 76633
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(293)-\sqrt{76633}}{2*-72}=\frac{-293-\sqrt{76633}}{-144}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(293)+\sqrt{76633}}{2*-72}=\frac{-293+\sqrt{76633}}{-144}
$