(4x-5)(10x-3)=(97-2x)

Solution for (4x-5)(10x-3)=(97-2x) equation:

(4x-5)(10x-3)=(97-2x)

We move all terms to the left:

(4x-5)(10x-3)-((97-2x))=0

We add all the numbers together, and all the variables

(4x-5)(10x-3)-((-2x+97))=0

We multiply parentheses ..

(+40x^2-12x-50x+15)-((-2x+97))=0


We calculate terms in parentheses: -((-2x+97)), so:

(-2x+97)

We get rid of parentheses

-2x+97

Back to the equation:

-(-2x+97)


We get rid of parentheses

40x^2-12x-50x+2x+15-97=0

We add all the numbers together, and all the variables

40x^2-60x-82=0

a = 40; b = -60; c = -82;
Δ = b2-4ac
Δ = -602-4·40·(-82)
Δ = 16720
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:

$\sqrt{\Delta}=\sqrt{16720}=\sqrt{16*1045}=\sqrt{16}*\sqrt{1045}=4\sqrt{1045}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{1045}}{2*40}=\frac{60-4\sqrt{1045}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{1045}}{2*40}=\frac{60+4\sqrt{1045}}{80}
$