16t^2-70t+65=0

Solution for 16t^2-70t+65=0 equation:

16t^2-70t+65=0

a = 16; b = -70; c = +65;
Δ = b2-4ac
Δ = -702-4·16·65
Δ = 740
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:

$\sqrt{\Delta}=\sqrt{740}=\sqrt{4*185}=\sqrt{4}*\sqrt{185}=2\sqrt{185}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-2\sqrt{185}}{2*16}=\frac{70-2\sqrt{185}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+2\sqrt{185}}{2*16}=\frac{70+2\sqrt{185}}{32}
$