9t^2+21t=54

Solution for 9t^2+21t=54 equation:

9t^2+21t=54

We move all terms to the left:

9t^2+21t-(54)=0

a = 9; b = 21; c = -54;
Δ = b2-4ac
Δ = 212-4·9·(-54)
Δ = 2385
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:

$\sqrt{\Delta}=\sqrt{2385}=\sqrt{9*265}=\sqrt{9}*\sqrt{265}=3\sqrt{265}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{265}}{2*9}=\frac{-21-3\sqrt{265}}{18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{265}}{2*9}=\frac{-21+3\sqrt{265}}{18}
$