t^2+5t=1

Solution for t^2+5t=1 equation:

t^2+5t=1

We move all terms to the left:

t^2+5t-(1)=0

a = 1; b = 5; c = -1;
Δ = b2-4ac
Δ = 52-4·1·(-1)
Δ = 29
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{29}}{2*1}=\frac{-5-\sqrt{29}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{29}}{2*1}=\frac{-5+\sqrt{29}}{2}
$