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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

-((+4x^2+10x-6x-15))+4x-x-1=0


We calculate terms in parentheses: -((+4x^2+10x-6x-15)), so:

(+4x^2+10x-6x-15)

We get rid of parentheses

4x^2+10x-6x-15

We add all the numbers together, and all the variables

4x^2+4x-15

Back to the equation:

-(4x^2+4x-15)


We add all the numbers together, and all the variables

3x-(4x^2+4x-15)-1=0

We get rid of parentheses

-4x^2+3x-4x+15-1=0

We add all the numbers together, and all the variables

-4x^2-1x+14=0

a = -4; b = -1; c = +14;
Δ = b2-4ac
Δ = -12-4·(-4)·14
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*-4}=\frac{-14}{-8}
=1+3/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*-4}=\frac{16}{-8}
=-2
$