0=t^2+3t-10

Solution for 0=t^2+3t-10 equation:

0=t^2+3t-10

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-t^2-3t+10=0

We add all the numbers together, and all the variables

-1t^2-3t+10=0

a = -1; b = -3; c = +10;
Δ = b2-4ac
Δ = -32-4·(-1)·10
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-1}=\frac{-4}{-2}
=+2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-1}=\frac{10}{-2}
=-5
$