X+1=t^2+5t

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

X+1=X^2+5X

We move all terms to the left:

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

We get rid of parentheses

-X^2+X-5X+1=0

We add all the numbers together, and all the variables

-1X^2-4X+1=0

a = -1; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·(-1)·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{5}}{2*-1}=\frac{4-2\sqrt{5}}{-2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{5}}{2*-1}=\frac{4+2\sqrt{5}}{-2}
$