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

Solution for 0=2(x^2-2x-3) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


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

2(x^2-2x-3)

We multiply parentheses

2x^2-4x-6

Back to the equation:

-(2x^2-4x-6)


We get rid of parentheses

-2x^2+4x+6=0

a = -2; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-2)·6
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-2}=\frac{4}{-4}
=-1
$