(7x-2+)(6x+4)=10

Solution for (7x-2+)(6x+4)=10 equation:

(7x-2+)(6x+4)=10

We move all terms to the left:

(7x-2+)(6x+4)-(10)=0

We add all the numbers together, and all the variables

(+7x)(6x+4)-10=0

We multiply parentheses ..

(+42x^2+28x)-10=0

We get rid of parentheses

42x^2+28x-10=0

a = 42; b = 28; c = -10;
Δ = b2-4ac
Δ = 282-4·42·(-10)
Δ = 2464
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:

$\sqrt{\Delta}=\sqrt{2464}=\sqrt{16*154}=\sqrt{16}*\sqrt{154}=4\sqrt{154}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{154}}{2*42}=\frac{-28-4\sqrt{154}}{84}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{154}}{2*42}=\frac{-28+4\sqrt{154}}{84}
$