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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-7=0

a = 2; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·2·(-7)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{14}}{2*2}=\frac{0-2\sqrt{14}}{4}
=-\frac{2\sqrt{14}}{4}
=-\frac{\sqrt{14}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{14}}{2*2}=\frac{0+2\sqrt{14}}{4}
=\frac{2\sqrt{14}}{4}
=\frac{\sqrt{14}}{2}
$