12x+21=3(4x+6)5x-20=3(x-4)

Solution for 12x+21=3(4x+6)5x-20=3(x-4) equation:

12x+21=3(4x+6)5x-20=3(x-4)

We move all terms to the left:

12x+21-(3(4x+6)5x-20)=0


We calculate terms in parentheses: -(3(4x+6)5x-20), so:

3(4x+6)5x-20

We multiply parentheses

60x^2+90x-20

Back to the equation:

-(60x^2+90x-20)


We get rid of parentheses

-60x^2+12x-90x+20+21=0

We add all the numbers together, and all the variables

-60x^2-78x+41=0

a = -60; b = -78; c = +41;
Δ = b2-4ac
Δ = -782-4·(-60)·41
Δ = 15924
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:

$\sqrt{\Delta}=\sqrt{15924}=\sqrt{4*3981}=\sqrt{4}*\sqrt{3981}=2\sqrt{3981}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-78)-2\sqrt{3981}}{2*-60}=\frac{78-2\sqrt{3981}}{-120}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-78)+2\sqrt{3981}}{2*-60}=\frac{78+2\sqrt{3981}}{-120}
$