(8x-3)(2x-5)=(4x-7)

Solution for (8x-3)(2x-5)=(4x-7) equation:

(8x-3)(2x-5)=(4x-7)

We move all terms to the left:

(8x-3)(2x-5)-((4x-7))=0

We multiply parentheses ..

(+16x^2-40x-6x+15)-((4x-7))=0


We calculate terms in parentheses: -((4x-7)), so:

(4x-7)

We get rid of parentheses

4x-7

Back to the equation:

-(4x-7)


We get rid of parentheses

16x^2-40x-6x-4x+15+7=0

We add all the numbers together, and all the variables

16x^2-50x+22=0

a = 16; b = -50; c = +22;
Δ = b2-4ac
Δ = -502-4·16·22
Δ = 1092
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:

$\sqrt{\Delta}=\sqrt{1092}=\sqrt{4*273}=\sqrt{4}*\sqrt{273}=2\sqrt{273}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{273}}{2*16}=\frac{50-2\sqrt{273}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{273}}{2*16}=\frac{50+2\sqrt{273}}{32}
$