-2t^2+8t+25=0

Solution for -2t^2+8t+25=0 equation:

-2t^2+8t+25=0

a = -2; b = 8; c = +25;
Δ = b2-4ac
Δ = 82-4·(-2)·25
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{66}}{2*-2}=\frac{-8-2\sqrt{66}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{66}}{2*-2}=\frac{-8+2\sqrt{66}}{-4}
$