4t^2+19t=5

Solution for 4t^2+19t=5 equation:

4t^2+19t=5

We move all terms to the left:

4t^2+19t-(5)=0

a = 4; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·4·(-5)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*4}=\frac{-40}{8}
=-5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*4}=\frac{2}{8}
=1/4
$