4t^2+5t-26=0

Solution for 4t^2+5t-26=0 equation:

4t^2+5t-26=0

a = 4; b = 5; c = -26;
Δ = b2-4ac
Δ = 52-4·4·(-26)
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{-(5)-21}{2*4}=\frac{-26}{8}
=-3+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+21}{2*4}=\frac{16}{8}
=2
$