(2x-17)(4x-18)=(8x-12)(2x-5)

Solution for (2x-17)(4x-18)=(8x-12)(2x-5) equation:

(2x-17)(4x-18)=(8x-12)(2x-5)

We move all terms to the left:

(2x-17)(4x-18)-((8x-12)(2x-5))=0

We multiply parentheses ..

(+8x^2-36x-68x+306)-((8x-12)(2x-5))=0


We calculate terms in parentheses: -((8x-12)(2x-5)), so:

(8x-12)(2x-5)

We multiply parentheses ..

(+16x^2-40x-24x+60)

We get rid of parentheses

16x^2-40x-24x+60

We add all the numbers together, and all the variables

16x^2-64x+60

Back to the equation:

-(16x^2-64x+60)


We get rid of parentheses

8x^2-16x^2-36x-68x+64x+306-60=0

We add all the numbers together, and all the variables

-8x^2-40x+246=0

a = -8; b = -40; c = +246;
Δ = b2-4ac
Δ = -402-4·(-8)·246
Δ = 9472
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:

$\sqrt{\Delta}=\sqrt{9472}=\sqrt{256*37}=\sqrt{256}*\sqrt{37}=16\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{37}}{2*-8}=\frac{40-16\sqrt{37}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{37}}{2*-8}=\frac{40+16\sqrt{37}}{-16}
$