6t^2+28t+27=0

Solution for 6t^2+28t+27=0 equation:

6t^2+28t+27=0

a = 6; b = 28; c = +27;
Δ = b2-4ac
Δ = 282-4·6·27
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{34}}{2*6}=\frac{-28-2\sqrt{34}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{34}}{2*6}=\frac{-28+2\sqrt{34}}{12}
$