5t=5t^2+4t-5

Solution for 5t=5t^2+4t-5 equation:

5t=5t^2+4t-5

We move all terms to the left:

5t-(5t^2+4t-5)=0

We get rid of parentheses

-5t^2+5t-4t+5=0

We add all the numbers together, and all the variables

-5t^2+t+5=0

a = -5; b = 1; c = +5;
Δ = b2-4ac
Δ = 12-4·(-5)·5
Δ = 101
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{-(1)-\sqrt{101}}{2*-5}=\frac{-1-\sqrt{101}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{101}}{2*-5}=\frac{-1+\sqrt{101}}{-10}
$