2t+6=2t(t+3)

Solution for 2t+6=2t(t+3) equation:

2t+6=2t(t+3)

We move all terms to the left:

2t+6-(2t(t+3))=0


We calculate terms in parentheses: -(2t(t+3)), so:

2t(t+3)

We multiply parentheses

2t^2+6t

Back to the equation:

-(2t^2+6t)


We get rid of parentheses

-2t^2+2t-6t+6=0

We add all the numbers together, and all the variables

-2t^2-4t+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:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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