(2x+3)(6+7x)=149

Solution for (2x+3)(6+7x)=149 equation:

(2x+3)(6+7x)=149

We move all terms to the left:

(2x+3)(6+7x)-(149)=0

We add all the numbers together, and all the variables

(2x+3)(7x+6)-149=0

We multiply parentheses ..

(+14x^2+12x+21x+18)-149=0

We get rid of parentheses

14x^2+12x+21x+18-149=0

We add all the numbers together, and all the variables

14x^2+33x-131=0

a = 14; b = 33; c = -131;
Δ = b2-4ac
Δ = 332-4·14·(-131)
Δ = 8425
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:

$\sqrt{\Delta}=\sqrt{8425}=\sqrt{25*337}=\sqrt{25}*\sqrt{337}=5\sqrt{337}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-5\sqrt{337}}{2*14}=\frac{-33-5\sqrt{337}}{28}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+5\sqrt{337}}{2*14}=\frac{-33+5\sqrt{337}}{28}
$