3t^2+10t=168

Solution for 3t^2+10t=168 equation:

3t^2+10t=168

We move all terms to the left:

3t^2+10t-(168)=0

a = 3; b = 10; c = -168;
Δ = b2-4ac
Δ = 102-4·3·(-168)
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-46}{2*3}=\frac{-56}{6}
=-9+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+46}{2*3}=\frac{36}{6}
=6
$