(2t^2-3t-7)=(4t^2-6t+1)

Solution for (2t^2-3t-7)=(4t^2-6t+1) equation:

(2t^2-3t-7)=(4t^2-6t+1)

We move all terms to the left:

(2t^2-3t-7)-((4t^2-6t+1))=0

We get rid of parentheses

2t^2-3t-((4t^2-6t+1))-7=0


We calculate terms in parentheses: -((4t^2-6t+1)), so:

(4t^2-6t+1)

We get rid of parentheses

4t^2-6t+1

Back to the equation:

-(4t^2-6t+1)


We get rid of parentheses

2t^2-4t^2-3t+6t-1-7=0

We add all the numbers together, and all the variables

-2t^2+3t-8=0

a = -2; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·(-2)·(-8)
Δ = -55
Delta is less than zero, so there is no solution for the equation