-3(t^2-4t-7)=0

Solution for -3(t^2-4t-7)=0 equation:

-3(t^2-4t-7)=0

We multiply parentheses

-3t^2+12t+21=0

a = -3; b = 12; c = +21;
Δ = b2-4ac
Δ = 122-4·(-3)·21
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{11}}{2*-3}=\frac{-12-6\sqrt{11}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{11}}{2*-3}=\frac{-12+6\sqrt{11}}{-6}
$