t2+15t+-200=0

Solution for t2+15t+-200=0 equation:

t2+15t+-200=0

We add all the numbers together, and all the variables

t^2+15t=0

a = 1; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·1·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*1}=\frac{-30}{2}
=-15
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*1}=\frac{0}{2}
=0
$