(2x+4)(3x-13)=116=62

Solution for (2x+4)(3x-13)=116=62 equation:

(2x+4)(3x-13)=116=62

We move all terms to the left:

(2x+4)(3x-13)-(116)=0

We multiply parentheses ..

(+6x^2-26x+12x-52)-116=0

We get rid of parentheses

6x^2-26x+12x-52-116=0

We add all the numbers together, and all the variables

6x^2-14x-168=0

a = 6; b = -14; c = -168;
Δ = b2-4ac
Δ = -142-4·6·(-168)
Δ = 4228
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:

$\sqrt{\Delta}=\sqrt{4228}=\sqrt{4*1057}=\sqrt{4}*\sqrt{1057}=2\sqrt{1057}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{1057}}{2*6}=\frac{14-2\sqrt{1057}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{1057}}{2*6}=\frac{14+2\sqrt{1057}}{12}
$