-t^2+9t+10=0

Solution for -t^2+9t+10=0 equation:

-t^2+9t+10=0

We add all the numbers together, and all the variables

-1t^2+9t+10=0

a = -1; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·(-1)·10
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*-1}=\frac{-20}{-2}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*-1}=\frac{2}{-2}
=-1
$