t2+45t-500=0

Solution for t2+45t-500=0 equation:

t2+45t-500=0

We add all the numbers together, and all the variables

t^2+45t-500=0

a = 1; b = 45; c = -500;
Δ = b2-4ac
Δ = 452-4·1·(-500)
Δ = 4025
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:

$\sqrt{\Delta}=\sqrt{4025}=\sqrt{25*161}=\sqrt{25}*\sqrt{161}=5\sqrt{161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-5\sqrt{161}}{2*1}=\frac{-45-5\sqrt{161}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+5\sqrt{161}}{2*1}=\frac{-45+5\sqrt{161}}{2}
$