(3x-4)(x+2)-5=0

Solution for (3x-4)(x+2)-5=0 equation:

(3x-4)(x+2)-5=0

We multiply parentheses ..

(+3x^2+6x-4x-8)-5=0

We get rid of parentheses

3x^2+6x-4x-8-5=0

We add all the numbers together, and all the variables

3x^2+2x-13=0

a = 3; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·3·(-13)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{10}}{2*3}=\frac{-2-4\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{10}}{2*3}=\frac{-2+4\sqrt{10}}{6}
$