5t2+-26t=0

Solution for 5t2+-26t=0 equation:

5t^2+-26t=0

We add all the numbers together, and all the variables

5t^2-26t=0

a = 5; b = -26; c = 0;
Δ = b2-4ac
Δ = -262-4·5·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-26}{2*5}=\frac{0}{10}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+26}{2*5}=\frac{52}{10}
=5+1/5
$