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

Solution for -4t^2+4t+3=0 equation:

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

a = -4; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·(-4)·3
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-4}=\frac{-12}{-8}
=1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-4}=\frac{4}{-8}
=-1/2
$