1=(6-3t)t

Solution for 1=(6-3t)t equation:

1=(6-3t)t

We move all terms to the left:

1-((6-3t)t)=0

We add all the numbers together, and all the variables

-((-3t+6)t)+1=0


We calculate terms in parentheses: -((-3t+6)t), so:

(-3t+6)t

We multiply parentheses

-3t^2+6t

Back to the equation:

-(-3t^2+6t)


We get rid of parentheses

3t^2-6t+1=0

a = 3; b = -6; c = +1;
Δ = b2-4ac
Δ = -62-4·3·1
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{6}}{2*3}=\frac{6-2\sqrt{6}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{6}}{2*3}=\frac{6+2\sqrt{6}}{6}
$