(2x+4)(3x+5)=50

Solution for (2x+4)(3x+5)=50 equation:

(2x+4)(3x+5)=50

We move all terms to the left:

(2x+4)(3x+5)-(50)=0

We multiply parentheses ..

(+6x^2+10x+12x+20)-50=0

We get rid of parentheses

6x^2+10x+12x+20-50=0

We add all the numbers together, and all the variables

6x^2+22x-30=0

a = 6; b = 22; c = -30;
Δ = b2-4ac
Δ = 222-4·6·(-30)
Δ = 1204
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:

$\sqrt{\Delta}=\sqrt{1204}=\sqrt{4*301}=\sqrt{4}*\sqrt{301}=2\sqrt{301}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{301}}{2*6}=\frac{-22-2\sqrt{301}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{301}}{2*6}=\frac{-22+2\sqrt{301}}{12}
$