t2+5t-5=0

Solution for t2+5t-5=0 equation:

t2+5t-5=0

We add all the numbers together, and all the variables

t^2+5t-5=0

a = 1; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·1·(-5)
Δ = 45
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{5}}{2*1}=\frac{-5-3\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{5}}{2*1}=\frac{-5+3\sqrt{5}}{2}
$