t(2t+1)=136

Solution for t(2t+1)=136 equation:

t(2t+1)=136

We move all terms to the left:

t(2t+1)-(136)=0

We multiply parentheses

2t^2+t-136=0

a = 2; b = 1; c = -136;
Δ = b2-4ac
Δ = 12-4·2·(-136)
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-33}{2*2}=\frac{-34}{4}
=-8+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+33}{2*2}=\frac{32}{4}
=8
$