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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

t^2-7t-2=0

a = 1; b = -7; c = -2;
Δ = b2-4ac
Δ = -72-4·1·(-2)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{57}}{2*1}=\frac{7-\sqrt{57}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{57}}{2*1}=\frac{7+\sqrt{57}}{2}
$