(3x-4)(x)=32

Solution for (3x-4)(x)=32 equation:

(3x-4)(x)=32

We move all terms to the left:

(3x-4)(x)-(32)=0

We multiply parentheses

3x^2-4x-32=0

a = 3; b = -4; c = -32;
Δ = b2-4ac
Δ = -42-4·3·(-32)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*3}=\frac{-16}{6}
=-2+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*3}=\frac{24}{6}
=4
$