5t+3t^2=168

Solution for 5t+3t^2=168 equation:

5t+3t^2=168

We move all terms to the left:

5t+3t^2-(168)=0

a = 3; b = 5; c = -168;
Δ = b2-4ac
Δ = 52-4·3·(-168)
Δ = 2041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{2041}}{2*3}=\frac{-5-\sqrt{2041}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{2041}}{2*3}=\frac{-5+\sqrt{2041}}{6}
$