-t^2+4t+6=0

Solution for -t^2+4t+6=0 equation:

-t^2+4t+6=0

We add all the numbers together, and all the variables

-1t^2+4t+6=0

a = -1; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-1)·6
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*-1}=\frac{-4-2\sqrt{10}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*-1}=\frac{-4+2\sqrt{10}}{-2}
$