7=((2x-3)*4x)

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

7=((2x-3)*4x)

We move all terms to the left:

7-(((2x-3)*4x))=0


We calculate terms in parentheses: -(((2x-3)*4x)), so:

((2x-3)*4x)


We calculate terms in parentheses: +((2x-3)*4x), so:

(2x-3)*4x

We multiply parentheses

8x^2-12x

Back to the equation:

+(8x^2-12x)


We get rid of parentheses

8x^2-12x

Back to the equation:

-(8x^2-12x)


We get rid of parentheses

-8x^2+12x+7=0

a = -8; b = 12; c = +7;
Δ = b2-4ac
Δ = 122-4·(-8)·7
Δ = 368
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:

$\sqrt{\Delta}=\sqrt{368}=\sqrt{16*23}=\sqrt{16}*\sqrt{23}=4\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{23}}{2*-8}=\frac{-12-4\sqrt{23}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{23}}{2*-8}=\frac{-12+4\sqrt{23}}{-16}
$