(4*x+7)*(2*x-3)=6*x^2

Solution for (4*x+7)*(2*x-3)=6*x^2 equation:

(4x+7)(2x-3)=6x^2

We move all terms to the left:

(4x+7)(2x-3)-(6x^2)=0

determiningTheFunctionDomain
-6x^2+(4x+7)(2x-3)=0

We multiply parentheses ..

-6x^2+(+8x^2-12x+14x-21)=0

We get rid of parentheses

-6x^2+8x^2-12x+14x-21=0

We add all the numbers together, and all the variables

2x^2+2x-21=0

a = 2; b = 2; c = -21;
Δ = b2-4ac
Δ = 22-4·2·(-21)
Δ = 172
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:

$\sqrt{\Delta}=\sqrt{172}=\sqrt{4*43}=\sqrt{4}*\sqrt{43}=2\sqrt{43}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{43}}{2*2}=\frac{-2-2\sqrt{43}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{43}}{2*2}=\frac{-2+2\sqrt{43}}{4}
$