12=(5-3x)(x+7)

Solution for 12=(5-3x)(x+7) equation:

12=(5-3x)(x+7)

We move all terms to the left:

12-((5-3x)(x+7))=0

We add all the numbers together, and all the variables

-((-3x+5)(x+7))+12=0

We multiply parentheses ..

-((-3x^2-21x+5x+35))+12=0


We calculate terms in parentheses: -((-3x^2-21x+5x+35)), so:

(-3x^2-21x+5x+35)

We get rid of parentheses

-3x^2-21x+5x+35

We add all the numbers together, and all the variables

-3x^2-16x+35

Back to the equation:

-(-3x^2-16x+35)


We get rid of parentheses

3x^2+16x-35+12=0

We add all the numbers together, and all the variables

3x^2+16x-23=0

a = 3; b = 16; c = -23;
Δ = b2-4ac
Δ = 162-4·3·(-23)
Δ = 532
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:

$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{133}}{2*3}=\frac{-16-2\sqrt{133}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{133}}{2*3}=\frac{-16+2\sqrt{133}}{6}
$