2t-5+4t^2=0

Solution for 2t-5+4t^2=0 equation:

2t-5+4t^2=0

a = 4; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·4·(-5)
Δ = 84
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*4}=\frac{-2-2\sqrt{21}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*4}=\frac{-2+2\sqrt{21}}{8}
$