(3x-1)(x+4)=56

Solution for (3x-1)(x+4)=56 equation:

(3x-1)(x+4)=56

We move all terms to the left:

(3x-1)(x+4)-(56)=0

We multiply parentheses ..

(+3x^2+12x-1x-4)-56=0

We get rid of parentheses

3x^2+12x-1x-4-56=0

We add all the numbers together, and all the variables

3x^2+11x-60=0

a = 3; b = 11; c = -60;
Δ = b2-4ac
Δ = 112-4·3·(-60)
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-29}{2*3}=\frac{-40}{6}
=-6+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+29}{2*3}=\frac{18}{6}
=3
$