(x-3)(x-4)=(2x+5)(x-1)

Solution for (x-3)(x-4)=(2x+5)(x-1) equation:

(x-3)(x-4)=(2x+5)(x-1)

We move all terms to the left:

(x-3)(x-4)-((2x+5)(x-1))=0

We multiply parentheses ..

(+x^2-4x-3x+12)-((2x+5)(x-1))=0


We calculate terms in parentheses: -((2x+5)(x-1)), so:

(2x+5)(x-1)

We multiply parentheses ..

(+2x^2-2x+5x-5)

We get rid of parentheses

2x^2-2x+5x-5

We add all the numbers together, and all the variables

2x^2+3x-5

Back to the equation:

-(2x^2+3x-5)


We get rid of parentheses

x^2-2x^2-4x-3x-3x+12+5=0

We add all the numbers together, and all the variables

-1x^2-10x+17=0

a = -1; b = -10; c = +17;
Δ = b2-4ac
Δ = -102-4·(-1)·17
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{42}}{2*-1}=\frac{10-2\sqrt{42}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{42}}{2*-1}=\frac{10+2\sqrt{42}}{-2}
$