-3t^2+18t=-4

Solution for -3t^2+18t=-4 equation:

-3t^2+18t=-4

We move all terms to the left:

-3t^2+18t-(-4)=0

We add all the numbers together, and all the variables

-3t^2+18t+4=0

a = -3; b = 18; c = +4;
Δ = b2-4ac
Δ = 182-4·(-3)·4
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{93}}{2*-3}=\frac{-18-2\sqrt{93}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{93}}{2*-3}=\frac{-18+2\sqrt{93}}{-6}
$