2t^2-6t+3=t^2+4t

Solution for 2t^2-6t+3=t^2+4t equation:

2t^2-6t+3=t^2+4t

We move all terms to the left:

2t^2-6t+3-(t^2+4t)=0

We get rid of parentheses

2t^2-t^2-6t-4t+3=0

We add all the numbers together, and all the variables

t^2-10t+3=0

a = 1; b = -10; c = +3;
Δ = b2-4ac
Δ = -102-4·1·3
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{22}}{2*1}=\frac{10-2\sqrt{22}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{22}}{2*1}=\frac{10+2\sqrt{22}}{2}
$