5t2=-9t+18

Solution for 5t2=-9t+18 equation:

5t^2=-9t+18

We move all terms to the left:

5t^2-(-9t+18)=0

We get rid of parentheses

5t^2+9t-18=0

a = 5; b = 9; c = -18;
Δ = b2-4ac
Δ = 92-4·5·(-18)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-21}{2*5}=\frac{-30}{10}
=-3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+21}{2*5}=\frac{12}{10}
=1+1/5
$