V(t)=t^2+10t-5

Solution for V(t)=t^2+10t-5 equation:

(V)=V^2+10V-5

We move all terms to the left:

(V)-(V^2+10V-5)=0

We get rid of parentheses

-V^2+V-10V+5=0

We add all the numbers together, and all the variables

-1V^2-9V+5=0

a = -1; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·(-1)·5
Δ = 101
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{101}}{2*-1}=\frac{9-\sqrt{101}}{-2}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{101}}{2*-1}=\frac{9+\sqrt{101}}{-2}
$