4t^2+3t+3=4

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

4t^2+3t+3=4

We move all terms to the left:

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

We add all the numbers together, and all the variables

4t^2+3t-1=0

a = 4; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·4·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*4}=\frac{-8}{8}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*4}=\frac{2}{8}
=1/4
$