(6x+4)(4x+3)=73

Solution for (6x+4)(4x+3)=73 equation:

(6x+4)(4x+3)=73

We move all terms to the left:

(6x+4)(4x+3)-(73)=0

We multiply parentheses ..

(+24x^2+18x+16x+12)-73=0

We get rid of parentheses

24x^2+18x+16x+12-73=0

We add all the numbers together, and all the variables

24x^2+34x-61=0

a = 24; b = 34; c = -61;
Δ = b2-4ac
Δ = 342-4·24·(-61)
Δ = 7012
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:

$\sqrt{\Delta}=\sqrt{7012}=\sqrt{4*1753}=\sqrt{4}*\sqrt{1753}=2\sqrt{1753}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{1753}}{2*24}=\frac{-34-2\sqrt{1753}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{1753}}{2*24}=\frac{-34+2\sqrt{1753}}{48}
$