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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4t^2-5t-1=0

a = -4; b = -5; c = -1;
Δ = b2-4ac
Δ = -52-4·(-4)·(-1)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*-4}=\frac{2}{-8}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*-4}=\frac{8}{-8}
=-1
$