-t^2+3t+10=10

Solution for -t^2+3t+10=10 equation:

-t^2+3t+10=10

We move all terms to the left:

-t^2+3t+10-(10)=0

We add all the numbers together, and all the variables

-1t^2+3t=0

a = -1; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-1)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-1}=\frac{-6}{-2}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-1}=\frac{0}{-2}
=0
$