(4x+2)(3x-7)=58

Solution for (4x+2)(3x-7)=58 equation:

(4x+2)(3x-7)=58

We move all terms to the left:

(4x+2)(3x-7)-(58)=0

We multiply parentheses ..

(+12x^2-28x+6x-14)-58=0

We get rid of parentheses

12x^2-28x+6x-14-58=0

We add all the numbers together, and all the variables

12x^2-22x-72=0

a = 12; b = -22; c = -72;
Δ = b2-4ac
Δ = -222-4·12·(-72)
Δ = 3940
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:

$\sqrt{\Delta}=\sqrt{3940}=\sqrt{4*985}=\sqrt{4}*\sqrt{985}=2\sqrt{985}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{985}}{2*12}=\frac{22-2\sqrt{985}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{985}}{2*12}=\frac{22+2\sqrt{985}}{24}
$