t(4t+13)=160

Solution for t(4t+13)=160 equation:

t(4t+13)=160

We move all terms to the left:

t(4t+13)-(160)=0

We multiply parentheses

4t^2+13t-160=0

a = 4; b = 13; c = -160;
Δ = b2-4ac
Δ = 132-4·4·(-160)
Δ = 2729
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{-(13)-\sqrt{2729}}{2*4}=\frac{-13-\sqrt{2729}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{2729}}{2*4}=\frac{-13+\sqrt{2729}}{8}
$