-5t^2+10t+14=0

Solution for -5t^2+10t+14=0 equation:

-5t^2+10t+14=0

a = -5; b = 10; c = +14;
Δ = b2-4ac
Δ = 102-4·(-5)·14
Δ = 380
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:

$\sqrt{\Delta}=\sqrt{380}=\sqrt{4*95}=\sqrt{4}*\sqrt{95}=2\sqrt{95}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{95}}{2*-5}=\frac{-10-2\sqrt{95}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{95}}{2*-5}=\frac{-10+2\sqrt{95}}{-10}
$