26t+16t^2-40=0

Solution for 26t+16t^2-40=0 equation:

26t+16t^2-40=0

a = 16; b = 26; c = -40;
Δ = b2-4ac
Δ = 262-4·16·(-40)
Δ = 3236
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:

$\sqrt{\Delta}=\sqrt{3236}=\sqrt{4*809}=\sqrt{4}*\sqrt{809}=2\sqrt{809}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{809}}{2*16}=\frac{-26-2\sqrt{809}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{809}}{2*16}=\frac{-26+2\sqrt{809}}{32}
$