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

Solution for 2x^+5x-12=(2x-3)(x+4) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

7x-((2x-3)(x+4))-12=0

We multiply parentheses ..

-((+2x^2+8x-3x-12))+7x-12=0


We calculate terms in parentheses: -((+2x^2+8x-3x-12)), so:

(+2x^2+8x-3x-12)

We get rid of parentheses

2x^2+8x-3x-12

We add all the numbers together, and all the variables

2x^2+5x-12

Back to the equation:

-(2x^2+5x-12)


We add all the numbers together, and all the variables

7x-(2x^2+5x-12)-12=0

We get rid of parentheses

-2x^2+7x-5x+12-12=0

We add all the numbers together, and all the variables

-2x^2+2x=0

a = -2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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