(8x+3)(2x-4)=(16x-36)

Solution for (8x+3)(2x-4)=(16x-36) equation:

(8x+3)(2x-4)=(16x-36)

We move all terms to the left:

(8x+3)(2x-4)-((16x-36))=0

We multiply parentheses ..

(+16x^2-32x+6x-12)-((16x-36))=0


We calculate terms in parentheses: -((16x-36)), so:

(16x-36)

We get rid of parentheses

16x-36

Back to the equation:

-(16x-36)


We get rid of parentheses

16x^2-32x+6x-16x-12+36=0

We add all the numbers together, and all the variables

16x^2-42x+24=0

a = 16; b = -42; c = +24;
Δ = b2-4ac
Δ = -422-4·16·24
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{57}}{2*16}=\frac{42-2\sqrt{57}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{57}}{2*16}=\frac{42+2\sqrt{57}}{32}
$