3+(2x-3)(4x+1)=10(x^2-x)-12x

Solution for 3+(2x-3)(4x+1)=10(x^2-x)-12x equation:

3+(2x-3)(4x+1)=10(x^2-x)-12x

We move all terms to the left:

3+(2x-3)(4x+1)-(10(x^2-x)-12x)=0

We multiply parentheses ..

(+8x^2+2x-12x-3)-(10(x^2-x)-12x)+3=0


We calculate terms in parentheses: -(10(x^2-x)-12x), so:

10(x^2-x)-12x

We add all the numbers together, and all the variables

-12x+10(x^2-x)

We multiply parentheses

10x^2-12x-10x

We add all the numbers together, and all the variables

10x^2-22x

Back to the equation:

-(10x^2-22x)


We get rid of parentheses

8x^2-10x^2+2x-12x+22x-3+3=0

We add all the numbers together, and all the variables

-2x^2+12x=0

a = -2; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-2)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-2}=\frac{-24}{-4}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-2}=\frac{0}{-4}
=0
$