2t^2=11t-13=0

Solution for 2t^2=11t-13=0 equation:

2t^2=11t-13=0

We move all terms to the left:

2t^2-(11t-13)=0

We get rid of parentheses

2t^2-11t+13=0

a = 2; b = -11; c = +13;
Δ = b2-4ac
Δ = -112-4·2·13
Δ = 17
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{-(-11)-\sqrt{17}}{2*2}=\frac{11-\sqrt{17}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{17}}{2*2}=\frac{11+\sqrt{17}}{4}
$