(t)=-5t^2+18t+3

Solution for (t)=-5t^2+18t+3 equation:

(t)=-5t^2+18t+3

We move all terms to the left:

(t)-(-5t^2+18t+3)=0

We get rid of parentheses

5t^2-18t+t-3=0

We add all the numbers together, and all the variables

5t^2-17t-3=0

a = 5; b = -17; c = -3;
Δ = b2-4ac
Δ = -172-4·5·(-3)
Δ = 349
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-\sqrt{349}}{2*5}=\frac{17-\sqrt{349}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{349}}{2*5}=\frac{17+\sqrt{349}}{10}
$