t2+26t=0

Solution for t2+26t=0 equation:

t2+26t=0

We add all the numbers together, and all the variables

t^2+26t=0

a = 1; b = 26; c = 0;
Δ = b2-4ac
Δ = 262-4·1·0
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*1}=\frac{-52}{2}
=-26
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+26}{2*1}=\frac{0}{2}
=0
$