-t2+6t+40=0

Solution for -t2+6t+40=0 equation:

-t2+6t+40=0

We add all the numbers together, and all the variables

-1t^2+6t+40=0

a = -1; b = 6; c = +40;
Δ = b2-4ac
Δ = 62-4·(-1)·40
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*-1}=\frac{-20}{-2}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*-1}=\frac{8}{-2}
=-4
$