2t(t-3)+2=2(2t-4)

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

2t(t-3)+2=2(2t-4)

We move all terms to the left:

2t(t-3)+2-(2(2t-4))=0

We multiply parentheses

2t^2-6t-(2(2t-4))+2=0


We calculate terms in parentheses: -(2(2t-4)), so:

2(2t-4)

We multiply parentheses

4t-8

Back to the equation:

-(4t-8)


We get rid of parentheses

2t^2-6t-4t+8+2=0

We add all the numbers together, and all the variables

2t^2-10t+10=0

a = 2; b = -10; c = +10;
Δ = b2-4ac
Δ = -102-4·2·10
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{5}}{2*2}=\frac{10-2\sqrt{5}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{5}}{2*2}=\frac{10+2\sqrt{5}}{4}
$