45t+22t-3t^2=0

Solution for 45t+22t-3t^2=0 equation:

45t+22t-3t^2=0

We add all the numbers together, and all the variables

-3t^2+67t=0

a = -3; b = 67; c = 0;
Δ = b2-4ac
Δ = 672-4·(-3)·0
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(67)-67}{2*-3}=\frac{-134}{-6}
=22+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(67)+67}{2*-3}=\frac{0}{-6}
=0
$