5t+0.25t^2=5

Solution for 5t+0.25t^2=5 equation:

5t+0.25t^2=5

We move all terms to the left:

5t+0.25t^2-(5)=0

a = 0.25; b = 5; c = -5;
Δ = b2-4ac
Δ = 52-4·0.25·(-5)
Δ = 30
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{30}}{2*0.25}=\frac{-5-\sqrt{30}}{0.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{30}}{2*0.25}=\frac{-5+\sqrt{30}}{0.5}
$