6t^2+168t+951=0

Solution for 6t^2+168t+951=0 equation:

6t^2+168t+951=0

a = 6; b = 168; c = +951;
Δ = b2-4ac
Δ = 1682-4·6·951
Δ = 5400
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:

$\sqrt{\Delta}=\sqrt{5400}=\sqrt{900*6}=\sqrt{900}*\sqrt{6}=30\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(168)-30\sqrt{6}}{2*6}=\frac{-168-30\sqrt{6}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(168)+30\sqrt{6}}{2*6}=\frac{-168+30\sqrt{6}}{12}
$