2t2-11t+-13=0

Solution for 2t2-11t+-13=0 equation:

2t^2-11t+-13=0

We add all the numbers together, and all the variables

2t^2-11t=0

a = 2; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·2·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*2}=\frac{0}{4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*2}=\frac{22}{4}
=5+1/2
$